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tan(1/x)<= tan(1/(x+1))

Solución

tan (\frac{1}{x}) \leq tan (\frac{1}{x + 1})

Solución

2 πn < x < \frac{2}{19 π} + 2 πn or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} + 2 πn \leq x < \frac{2}{17 π} + 2 πn or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} + 2 πn \leq x < \frac{2}{15 π} + 2 πn or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} + 2 πn \leq x < \frac{2}{13 π} + 2 πn or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} + 2 πn \leq x < \frac{2}{11 π} + 2 πn or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} + 2 πn \leq x < \frac{2}{9 π} + 2 πn or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} + 2 πn \leq x < \frac{2}{7 π} + 2 πn or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} + 2 πn \leq x < \frac{2}{5 π} + 2 πn or \frac{- π + π ^{2} + 2 π}{2 π} + 2 πn \leq x < \frac{2}{3 π} + 2 πn or \frac{π ^{2} + 4 π - π}{2 π} + 2 πn \leq x < \frac{2}{π} + 2 πn

Notación de intervalos

(2 πn, \frac{2}{19 π} + 2 πn) \cup [\frac{- 9 π + 81 π ^{2} + 36 π}{18 π} + 2 πn, \frac{2}{17 π} + 2 πn) \cup [\frac{- 2 π + 4 π ^{2} + 2 π}{4 π} + 2 πn, \frac{2}{15 π} + 2 πn) \cup [\frac{- 7 π + 49 π ^{2} + 28 π}{14 π} + 2 πn, \frac{2}{13 π} + 2 πn) \cup [\frac{- 3 π + 9 π ^{2} + 6 π}{6 π} + 2 πn, \frac{2}{11 π} + 2 πn) \cup [\frac{- 5 π + 25 π ^{2} + 20 π}{10 π} + 2 πn, \frac{2}{9 π} + 2 πn) \cup [\frac{- 2 π + 4 π ^{2} + 4 π}{4 π} + 2 πn, \frac{2}{7 π} + 2 πn) \cup [\frac{- 3 π + 9 π ^{2} + 12 π}{6 π} + 2 πn, \frac{2}{5 π} + 2 πn) \cup [\frac{- π + π ^{2} + 2 π}{2 π} + 2 πn, \frac{2}{3 π} + 2 πn) \cup [\frac{π ^{2} + 4 π - π}{2 π} + 2 πn, \frac{2}{π} + 2 πn)

Notación decimal

2 πn < x < 0.03350 \dots + 2 πn or 0.03419 \dots + 2 πn \leq x < 0.03744 \dots + 2 πn or 0.03832 \dots + 2 πn \leq x < 0.04244 \dots + 2 πn or 0.04357 \dots + 2 πn \leq x < 0.04897 \dots + 2 πn or 0.05050 \dots + 2 πn \leq x < 0.05787 \dots + 2 πn or 0.06005 \dots + 2 πn \leq x < 0.07073 \dots + 2 πn or 0.07408 \dots + 2 πn \leq x < 0.09094 \dots + 2 πn or 0.09674 \dots + 2 πn \leq x < 0.12732 \dots + 2 πn or 0.13965 \dots + 2 πn \leq x < 0.21220 \dots + 2 πn or 0.25386 \dots + 2 πn \leq x < 0.63661 \dots + 2 πn

Pasos de solución

tan (\frac{1}{x}) \leq tan (\frac{1}{x + 1})

Mover

tan (\frac{1}{x + 1})

al lado izquierdo

tan (\frac{1}{x}) \leq tan (\frac{1}{x + 1})

Restar

tan (\frac{1}{x + 1})

de ambos lados

tan (\frac{1}{x}) - tan (\frac{1}{x + 1}) \leq tan (\frac{1}{x + 1}) - tan (\frac{1}{x + 1})

tan (\frac{1}{x}) - tan (\frac{1}{x + 1}) \leq 0

tan (\frac{1}{x}) - tan (\frac{1}{x + 1}) \leq 0

Periodicidad de

tan (\frac{1}{x}) - tan (\frac{1}{x + 1}) :

No periodico

La función

tan (\frac{1}{x}) - tan (\frac{1}{x + 1})

no es periodica

= No periodico

Expresar con seno, coseno

tan (\frac{1}{x}) - tan (\frac{1}{x + 1}) \leq 0

Utilizar la identidad trigonométrica básica:

tan (x) = \frac{sin ( x )}{cos ( x )}

\frac{sin ( \frac{1}{x} )}{cos ( \frac{1}{x} )} - tan (\frac{1}{x + 1}) \leq 0

Utilizar la identidad trigonométrica básica:

tan (x) = \frac{sin ( x )}{cos ( x )}

\frac{sin ( \frac{1}{x} )}{cos ( \frac{1}{x} )} - \frac{sin ( \frac{1}{x + 1} )}{cos ( \frac{1}{x + 1} )} \leq 0

\frac{sin ( \frac{1}{x} )}{cos ( \frac{1}{x} )} - \frac{sin ( \frac{1}{x + 1} )}{cos ( \frac{1}{x + 1} )} \leq 0

Simplificar

\frac{sin ( \frac{1}{x} )}{cos ( \frac{1}{x} )} - \frac{sin ( \frac{1}{x + 1} )}{cos ( \frac{1}{x + 1} )} : \frac{sin ( \frac{1}{x} ) cos ( \frac{1}{x + 1} ) - sin ( \frac{1}{x + 1} ) cos ( \frac{1}{x} )}{cos ( \frac{1}{x} ) cos ( \frac{1}{x + 1} )}

\frac{sin ( \frac{1}{x} )}{cos ( \frac{1}{x} )} - \frac{sin ( \frac{1}{x + 1} )}{cos ( \frac{1}{x + 1} )}

Mínimo común múltiplo de

cos (\frac{1}{x}), cos (\frac{1}{x + 1}) : cos (\frac{1}{x}) cos (\frac{1}{x + 1})

cos (\frac{1}{x}), cos (\frac{1}{x + 1})

Mínimo común múltiplo (MCM)

Calcular una expresión que este compuesta de factores que aparezcan tanto en

cos (\frac{1}{x})

cos (\frac{1}{x + 1})

= cos (\frac{1}{x}) cos (\frac{1}{x + 1})

Reescribir las fracciones basandose en el mínimo común denominador

Multiplicar cada numerador por la misma cantidad necesaria para multiplicar el denominador correspondiente y convertirlo en el mínimo común denominador

Para

\frac{sin ( \frac{1}{x} )}{cos ( \frac{1}{x} )} :

multiplicar el denominador y el numerador por

cos (\frac{1}{x + 1})

\frac{sin ( \frac{1}{x} )}{cos ( \frac{1}{x} )} = \frac{sin ( \frac{1}{x} ) cos ( \frac{1}{x + 1} )}{cos ( \frac{1}{x} ) cos ( \frac{1}{x + 1} )}

Para

\frac{sin ( \frac{1}{x + 1} )}{cos ( \frac{1}{x + 1} )} :

multiplicar el denominador y el numerador por

cos (\frac{1}{x})

\frac{sin ( \frac{1}{x + 1} )}{cos ( \frac{1}{x + 1} )} = \frac{sin ( \frac{1}{x + 1} ) cos ( \frac{1}{x} )}{cos ( \frac{1}{x + 1} ) cos ( \frac{1}{x} )}

= \frac{sin ( \frac{1}{x} ) cos ( \frac{1}{x + 1} )}{cos ( \frac{1}{x} ) cos ( \frac{1}{x + 1} )} - \frac{sin ( \frac{1}{x + 1} ) cos ( \frac{1}{x} )}{cos ( \frac{1}{x + 1} ) cos ( \frac{1}{x} )}

Ya que los denominadores son iguales, combinar las fracciones:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ( \frac{1}{x} ) cos ( \frac{1}{x + 1} ) - sin ( \frac{1}{x + 1} ) cos ( \frac{1}{x} )}{cos ( \frac{1}{x} ) cos ( \frac{1}{x + 1} )}

\frac{sin ( \frac{1}{x} ) cos ( \frac{1}{x + 1} ) - sin ( \frac{1}{x + 1} ) cos ( \frac{1}{x} )}{cos ( \frac{1}{x} ) cos ( \frac{1}{x + 1} )} \leq 0

Encontrar los ceros y puntos indefinidos de

\frac{sin ( \frac{1}{x} ) cos ( \frac{1}{x + 1} ) - sin ( \frac{1}{x + 1} ) cos ( \frac{1}{x} )}{cos ( \frac{1}{x} ) cos ( \frac{1}{x + 1} )}

para

0 \leq x < 2 π

Para encontrar los ceros, transformar la desigualdad a 0

\frac{sin ( \frac{1}{x} ) cos ( \frac{1}{x + 1} ) - sin ( \frac{1}{x + 1} ) cos ( \frac{1}{x} )}{cos ( \frac{1}{x} ) cos ( \frac{1}{x + 1} )} = 0

\frac{sin ( \frac{1}{x} ) cos ( \frac{1}{x + 1} ) - sin ( \frac{1}{x + 1} ) cos ( \frac{1}{x} )}{cos ( \frac{1}{x} ) cos ( \frac{1}{x + 1} )} = 0, 0 \leq x < 2 π : x = \frac{π ^{2} + 4 π - π}{2 π}, x = \frac{- π + π ^{2} + 2 π}{2 π}, x = \frac{- 3 π + 9 π ^{2} + 12 π}{6 π}, x = \frac{- 2 π + 4 π ^{2} + 4 π}{4 π}, x = \frac{- 5 π + 25 π ^{2} + 20 π}{10 π}, x = \frac{- 3 π + 9 π ^{2} + 6 π}{6 π}, x = \frac{- 7 π + 49 π ^{2} + 28 π}{14 π}, x = \frac{- 2 π + 4 π ^{2} + 2 π}{4 π}, x = \frac{- 9 π + 81 π ^{2} + 36 π}{18 π}; n \neq = 0, n \neq = - \frac{1}{2}

\frac{sin ( \frac{1}{x} ) cos ( \frac{1}{x + 1} ) - sin ( \frac{1}{x + 1} ) cos ( \frac{1}{x} )}{cos ( \frac{1}{x} ) cos ( \frac{1}{x + 1} )} = 0, 0 \leq x < 2 π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (\frac{1}{x}) cos (\frac{1}{x + 1}) - sin (\frac{1}{x + 1}) cos (\frac{1}{x}) = 0

Re-escribir usando identidades trigonométricas

sin (\frac{1}{x}) cos (\frac{1}{x + 1}) - sin (\frac{1}{x + 1}) cos (\frac{1}{x})

Utilizar la identidad de diferencia de ángulos:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

= sin (\frac{1}{x} - \frac{1}{x + 1})

sin (\frac{1}{x} - \frac{1}{x + 1}) = 0

Soluciones generales para

sin (\frac{1}{x} - \frac{1}{x + 1}) = 0

tabla de valores periódicos con

2 πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

\frac{1}{x} - \frac{1}{x + 1} = 0 + 2 πn, \frac{1}{x} - \frac{1}{x + 1} = π + 2 πn

\frac{1}{x} - \frac{1}{x + 1} = 0 + 2 πn, \frac{1}{x} - \frac{1}{x + 1} = π + 2 πn

Resolver

\frac{1}{x} - \frac{1}{x + 1} = 0 + 2 πn : x = \frac{- πn + πn ( πn + 2 )}{2 πn}, x = - \frac{πn + πn ( πn + 2 )}{2 πn}; n \neq = 0

\frac{1}{x} - \frac{1}{x + 1} = 0 + 2 πn

Multiplicar por el mínimo común múltiplo

\frac{1}{x} - \frac{1}{x + 1} = 0 + 2 πn

Encontrar el mínimo común múltiplo de

x, x + 1 : x (x + 1)

x, x + 1

Mínimo común múltiplo (MCM)

Calcular una expresión que este compuesta de factores que aparezcan tanto en

x

x + 1

= x (x + 1)

Multiplicar por el mínimo común múltiplo=

x (x + 1)

\frac{1}{x} x (x + 1) - \frac{1}{x + 1} x (x + 1) = 0 \cdot x (x + 1) + 2 πn x (x + 1)

Simplificar

\frac{1}{x} x (x + 1) - \frac{1}{x + 1} x (x + 1) = 0 \cdot x (x + 1) + 2 πn x (x + 1)

Simplificar

\frac{1}{x} x (x + 1) : x + 1

\frac{1}{x} x (x + 1)

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot x ( x + 1 )}{x}

Eliminar los terminos comunes:

x

= 1 \cdot (x + 1)

Simplificar

= x + 1

Simplificar

- \frac{1}{x + 1} x (x + 1) : - x

- \frac{1}{x + 1} x (x + 1)

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{1 \cdot x ( x + 1 )}{x + 1}

Eliminar los terminos comunes:

x + 1

= - 1 \cdot x

Multiplicar:

1 \cdot x = x

= - x

Simplificar

0 \cdot x (x + 1) : 0

0 \cdot x (x + 1)

Aplicar la regla

0 \cdot a = 0

= 0

x + 1 - x = 0 + 2 πn x (x + 1)

x + 1 - x = 1

x + 1 - x

Agrupar términos semejantes

= x - x + 1

Sumar elementos similares:

x - x = 0

= 1

Simplificar

0 + 2 πn x (x + 1) : 2 πn x (x + 1)

0 + 2 πn x (x + 1)

0 + 2 πn x (x + 1) = 2 πn x (x + 1)

= 2 πn x (x + 1)

1 = 2 πn x (x + 1)

1 = 2 πn x (x + 1)

1 = 2 πn x (x + 1)

Resolver

1 = 2 πn x (x + 1) : x = \frac{- πn + πn ( πn + 2 )}{2 πn}, x = - \frac{πn + πn ( πn + 2 )}{2 πn}; n \neq = 0

1 = 2 πn x (x + 1)

Desarrollar

2 πn x (x + 1) : 2 πn x^{2} + 2 πn x

2 πn x (x + 1)

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = 2 πn x, b = x, c = 1

= 2 πn xx + 2 πn x \cdot 1

= 2 πn xx + 2 \cdot 1 πn x

Simplificar

2 πn xx + 2 \cdot 1 πn x : 2 πn x^{2} + 2 πn x

2 πn xx + 2 \cdot 1 πn x

2 πn xx = 2 πn x^{2}

2 πn xx

Aplicar las leyes de los exponentes:

a^{b} \cdot a^{c} = a^{b + c}

xx = x^{1 + 1}

= 2 πn x^{1 + 1}

Sumar:

1 + 1 = 2

= 2 πn x^{2}

2 \cdot 1 πn x = 2 πn x

2 \cdot 1 πn x

Multiplicar los numeros:

2 \cdot 1 = 2

= 2 πn x

= 2 πn x^{2} + 2 πn x

= 2 πn x^{2} + 2 πn x

1 = 2 πn x^{2} + 2 πn x

Intercambiar lados

2 πn x^{2} + 2 πn x = 1

Mover

1

al lado izquierdo

2 πn x^{2} + 2 πn x = 1

Restar

1

de ambos lados

2 πn x^{2} + 2 πn x - 1 = 1 - 1

Simplificar

2 πn x^{2} + 2 πn x - 1 = 0

2 πn x^{2} + 2 πn x - 1 = 0

Resolver con la fórmula general para ecuaciones de segundo grado:

2 πn x^{2} + 2 πn x - 1 = 0

Formula general para ecuaciones de segundo grado:

Para

a = 2 πn, b = 2 πn, c = - 1

x_{1, 2} = \frac{- 2 πn \pm ( 2 πn ) ^{2} - 4 \cdot 2 πn ( - 1 )}{2 \cdot 2 πn}

x_{1, 2} = \frac{- 2 πn \pm ( 2 πn ) ^{2} - 4 \cdot 2 πn ( - 1 )}{2 \cdot 2 πn}

Simplificar

(2 πn)^{2} - 4 \cdot 2 πn (- 1) : 2 πn (πn + 2)

(2 πn)^{2} - 4 \cdot 2 πn (- 1)

Aplicar la regla

- (- a) = a

= (2 πn)^{2} + 4 \cdot 2 πn \cdot 1

Aplicar las leyes de los exponentes:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} π^{2} n^{2} + 4 \cdot 2 \cdot 1 πn

Multiplicar los numeros:

4 \cdot 2 \cdot 1 = 8

= 2^{2} π^{2} n^{2} + 8 πn

Factorizar

2^{2} π^{2} n^{2} + 8 πn : 4 πn (πn + 2)

2^{2} π^{2} n^{2} + 8 πn

Aplicar las leyes de los exponentes:

a^{b + c} = a^{b} a^{c}

π^{2} = ππ, n^{2} = nn

= 4 ππnn + 8 πn

Reescribir como

= 4 πnπn + 2 \cdot 4 πn

Factorizar el termino común

4 πn

= 4 πn (πn + 2)

= 4 πn (πn + 2)

Aplicar la siguiente propiedad de los radicales:

asumiendo que

a \geq 0, b \geq 0

= 4 πn (πn + 2)

4 = 2

4

Descomponer el número en factores primos:

4 = 2^{2}

= 2^{2}

Aplicar las leyes de los exponentes:

2^{2} = 2

= 2

= 2 πn (πn + 2)

x_{1, 2} = \frac{- 2 πn \pm 2 πn ( πn + 2 )}{2 \cdot 2 πn}; n \neq = 0

Separar las soluciones

x_{1} = \frac{- 2 πn + 2 πn ( πn + 2 )}{2 \cdot 2 πn}, x_{2} = \frac{- 2 πn - 2 πn ( πn + 2 )}{2 \cdot 2 πn}

x = \frac{- 2 πn + 2 πn ( πn + 2 )}{2 \cdot 2 πn} : \frac{- πn + πn ( πn + 2 )}{2 πn}

\frac{- 2 πn + 2 πn ( πn + 2 )}{2 \cdot 2 πn}

Multiplicar los numeros:

2 \cdot 2 = 4

= \frac{- 2 πn + 2 πn ( πn + 2 )}{4 πn}

Factorizar el termino común

2

= \frac{2 ( - πn + ( 2 + nπ ) nπ )}{4 πn}

Eliminar los terminos comunes:

2

= \frac{- πn + πn ( πn + 2 )}{2 πn}

x = \frac{- 2 πn - 2 πn ( πn + 2 )}{2 \cdot 2 πn} : - \frac{πn + πn ( πn + 2 )}{2 πn}

\frac{- 2 πn - 2 πn ( πn + 2 )}{2 \cdot 2 πn}

Multiplicar los numeros:

2 \cdot 2 = 4

= \frac{- 2 πn - 2 πn ( πn + 2 )}{4 πn}

Factorizar el termino común

2

= - \frac{2 ( πn + ( 2 + nπ ) nπ )}{4 πn}

Eliminar los terminos comunes:

2

= - \frac{πn + πn ( πn + 2 )}{2 πn}

Las soluciones a la ecuación de segundo grado son:

x = \frac{- πn + πn ( πn + 2 )}{2 πn}, x = - \frac{πn + πn ( πn + 2 )}{2 πn}; n \neq = 0

x = \frac{- πn + πn ( πn + 2 )}{2 πn}, x = - \frac{πn + πn ( πn + 2 )}{2 πn}; n \neq = 0

Resolver

\frac{1}{x} - \frac{1}{x + 1} = π + 2 πn : x = \frac{- π - 2 πn + ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}, x = \frac{- π - 2 πn - ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}; n \neq = - \frac{1}{2}

\frac{1}{x} - \frac{1}{x + 1} = π + 2 πn

Multiplicar por el mínimo común múltiplo

\frac{1}{x} - \frac{1}{x + 1} = π + 2 πn

Encontrar el mínimo común múltiplo de

x, x + 1 : x (x + 1)

x, x + 1

Mínimo común múltiplo (MCM)

Calcular una expresión que este compuesta de factores que aparezcan tanto en

x

x + 1

= x (x + 1)

Multiplicar por el mínimo común múltiplo=

x (x + 1)

\frac{1}{x} x (x + 1) - \frac{1}{x + 1} x (x + 1) = π x (x + 1) + 2 πn x (x + 1)

Simplificar

\frac{1}{x} x (x + 1) - \frac{1}{x + 1} x (x + 1) = π x (x + 1) + 2 πn x (x + 1)

Simplificar

\frac{1}{x} x (x + 1) : x + 1

\frac{1}{x} x (x + 1)

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot x ( x + 1 )}{x}

Eliminar los terminos comunes:

x

= 1 \cdot (x + 1)

Simplificar

= x + 1

Simplificar

- \frac{1}{x + 1} x (x + 1) : - x

- \frac{1}{x + 1} x (x + 1)

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{1 \cdot x ( x + 1 )}{x + 1}

Eliminar los terminos comunes:

x + 1

= - 1 \cdot x

Multiplicar:

1 \cdot x = x

= - x

x + 1 - x = π x (x + 1) + 2 πn x (x + 1)

x + 1 - x = 1

x + 1 - x

Agrupar términos semejantes

= x - x + 1

Sumar elementos similares:

x - x = 0

= 1

1 = π x (x + 1) + 2 πn x (x + 1)

1 = π x (x + 1) + 2 πn x (x + 1)

1 = π x (x + 1) + 2 πn x (x + 1)

Resolver

1 = π x (x + 1) + 2 πn x (x + 1) : x = \frac{- π - 2 πn + ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}, x = \frac{- π - 2 πn - ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}; n \neq = - \frac{1}{2}

1 = π x (x + 1) + 2 πn x (x + 1)

Desarrollar

π x (x + 1) + 2 πn x (x + 1) : π x^{2} + π x + 2 πn x^{2} + 2 πn x

π x (x + 1) + 2 πn x (x + 1)

Expandir

π x (x + 1) : π x^{2} + π x

π x (x + 1)

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = π x, b = x, c = 1

= π xx + π x \cdot 1

= π xx + 1 π x

Simplificar

π xx + 1 π x : π x^{2} + π x

π xx + 1 π x

π xx = π x^{2}

π xx

Aplicar las leyes de los exponentes:

a^{b} \cdot a^{c} = a^{b + c}

xx = x^{1 + 1}

= π x^{1 + 1}

Sumar:

1 + 1 = 2

= π x^{2}

1 π x = π x

1 π x

Multiplicar:

1 π = π

= π x

= π x^{2} + π x

= π x^{2} + π x

= π x^{2} + π x + 2 πn x (x + 1)

Expandir

2 πn x (x + 1) : 2 πn x^{2} + 2 πn x

2 πn x (x + 1)

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = 2 πn x, b = x, c = 1

= 2 πn xx + 2 πn x \cdot 1

= 2 πn xx + 2 \cdot 1 πn x

Simplificar

2 πn xx + 2 \cdot 1 πn x : 2 πn x^{2} + 2 πn x

2 πn xx + 2 \cdot 1 πn x

2 πn xx = 2 πn x^{2}

2 πn xx

Aplicar las leyes de los exponentes:

a^{b} \cdot a^{c} = a^{b + c}

xx = x^{1 + 1}

= 2 πn x^{1 + 1}

Sumar:

1 + 1 = 2

= 2 πn x^{2}

2 \cdot 1 πn x = 2 πn x

2 \cdot 1 πn x

Multiplicar los numeros:

2 \cdot 1 = 2

= 2 πn x

= 2 πn x^{2} + 2 πn x

= 2 πn x^{2} + 2 πn x

= π x^{2} + π x + 2 πn x^{2} + 2 πn x

1 = π x^{2} + π x + 2 πn x^{2} + 2 πn x

Intercambiar lados

π x^{2} + π x + 2 πn x^{2} + 2 πn x = 1

Mover

1

al lado izquierdo

π x^{2} + π x + 2 πn x^{2} + 2 πn x = 1

Restar

1

de ambos lados

π x^{2} + π x + 2 πn x^{2} + 2 πn x - 1 = 1 - 1

Simplificar

π x^{2} + π x + 2 πn x^{2} + 2 πn x - 1 = 0

π x^{2} + π x + 2 πn x^{2} + 2 πn x - 1 = 0

Escribir en la forma binómica

a x^{2} + b x + c = 0

(π + 2 πn) x^{2} + (π + 2 πn) x - 1 = 0

Resolver con la fórmula general para ecuaciones de segundo grado:

(π + 2 πn) x^{2} + (π + 2 πn) x - 1 = 0

Formula general para ecuaciones de segundo grado:

Para

a = π + 2 πn, b = π + 2 πn, c = - 1

x_{1, 2} = \frac{- ( π + 2 πn ) \pm ( π + 2 πn ) ^{2} - 4 ( π + 2 πn ) ( - 1 )}{2 ( π + 2 πn )}

x_{1, 2} = \frac{- ( π + 2 πn ) \pm ( π + 2 πn ) ^{2} - 4 ( π + 2 πn ) ( - 1 )}{2 ( π + 2 πn )}

Simplificar

(π + 2 πn)^{2} - 4 (π + 2 πn) (- 1) : (π + 2 πn)^{2} + 4 (π + 2 πn)

(π + 2 πn)^{2} - 4 (π + 2 πn) (- 1)

Aplicar la regla

- (- a) = a

= (π + 2 πn)^{2} + 4 (π + 2 πn) \cdot 1

Multiplicar los numeros:

4 \cdot 1 = 4

= (π + 2 πn)^{2} + 4 (π + 2 πn)

x_{1, 2} = \frac{- ( π + 2 πn ) \pm ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}; n \neq = - \frac{1}{2}

Separar las soluciones

x_{1} = \frac{- ( π + 2 πn ) + ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}, x_{2} = \frac{- ( π + 2 πn ) - ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}

x = \frac{- ( π + 2 πn ) + ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )} : \frac{- π - 2 πn + ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}

\frac{- ( π + 2 πn ) + ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}

- (π + 2 πn) : - π - 2 πn

- (π + 2 πn)

Poner los parentesis

= - (π) - (2 πn)

Aplicar las reglas de los signos

+ (- a) = - a

= - π - 2 πn

= \frac{- π - 2 πn + ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}

x = \frac{- ( π + 2 πn ) - ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )} : \frac{- π - 2 πn - ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}

\frac{- ( π + 2 πn ) - ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}

- (π + 2 πn) : - π - 2 πn

- (π + 2 πn)

Poner los parentesis

= - (π) - (2 πn)

Aplicar las reglas de los signos

+ (- a) = - a

= - π - 2 πn

= \frac{- π - 2 πn - ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}

Las soluciones a la ecuación de segundo grado son:

x = \frac{- π - 2 πn + ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}, x = \frac{- π - 2 πn - ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}; n \neq = - \frac{1}{2}

x = \frac{- π - 2 πn + ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}, x = \frac{- π - 2 πn - ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}; n \neq = - \frac{1}{2}

x = \frac{- πn + πn ( πn + 2 )}{2 πn}, x = - \frac{πn + πn ( πn + 2 )}{2 πn}, x = \frac{- π - 2 πn + ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}, x = \frac{- π - 2 πn - ( π + 2 πn ) ^{2} + 4 ( π + 2 πn )}{2 ( π + 2 πn )}; n \neq = 0, n \neq = - \frac{1}{2}

Soluciones para el rango

0 \leq x < 2 π

x = \frac{π ^{2} + 4 π - π}{2 π}, x = \frac{- π + π ^{2} + 2 π}{2 π}, x = \frac{- 3 π + 9 π ^{2} + 12 π}{6 π}, x = \frac{- 2 π + 4 π ^{2} + 4 π}{4 π}, x = \frac{- 5 π + 25 π ^{2} + 20 π}{10 π}, x = \frac{- 3 π + 9 π ^{2} + 6 π}{6 π}, x = \frac{- 7 π + 49 π ^{2} + 28 π}{14 π}, x = \frac{- 2 π + 4 π ^{2} + 2 π}{4 π}, x = \frac{- 9 π + 81 π ^{2} + 36 π}{18 π}; n \neq = 0, n \neq = - \frac{1}{2}

Encontrar los puntos indefinidos:

x = \frac{2}{π}, x = \frac{2}{3 π}, x = \frac{2}{5 π}, x = \frac{2}{7 π}, x = \frac{2}{9 π}, x = \frac{2}{11 π}, x = \frac{2}{13 π}, x = \frac{2}{15 π}, x = \frac{2}{17 π}, x = \frac{2}{19 π}

Encontrar los ceros del denominador

cos (\frac{1}{x}) cos (\frac{1}{x + 1}) = 0

Resolver cada parte por separado

cos (\frac{1}{x}) = 0 or cos (\frac{1}{x + 1}) = 0

cos (\frac{1}{x}) = 0, 0 \leq x < 2 π : x = \frac{2}{π}, x = \frac{2}{3 π}, x = \frac{2}{5 π}, x = \frac{2}{7 π}, x = \frac{2}{9 π}, x = \frac{2}{11 π}, x = \frac{2}{13 π}, x = \frac{2}{15 π}, x = \frac{2}{17 π}, x = \frac{2}{19 π}

cos (\frac{1}{x}) = 0, 0 \leq x < 2 π

Soluciones generales para

cos (\frac{1}{x}) = 0

cos (x)

tabla de valores periódicos con

2 πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

\frac{1}{x} = \frac{π}{2} + 2 πn, \frac{1}{x} = \frac{3 π}{2} + 2 πn

\frac{1}{x} = \frac{π}{2} + 2 πn, \frac{1}{x} = \frac{3 π}{2} + 2 πn

Resolver

\frac{1}{x} = \frac{π}{2} + 2 πn : x = \frac{2}{π ( 1 + 4 n )}; n \neq = - \frac{1}{4}

\frac{1}{x} = \frac{π}{2} + 2 πn

Multiplicar ambos lados por

x

\frac{1}{x} = \frac{π}{2} + 2 πn

Multiplicar ambos lados por

x

\frac{1}{x} x = \frac{π}{2} x + 2 πn x

Simplificar

1 = \frac{π}{2} x + 2 πn x

1 = \frac{π}{2} x + 2 πn x

Intercambiar lados

\frac{π}{2} x + 2 πn x = 1

Multiplicar ambos lados por

2

\frac{π}{2} x + 2 πn x = 1

Multiplicar ambos lados por

2

\frac{π}{2} x \cdot 2 + 2 πn x \cdot 2 = 1 \cdot 2

Simplificar

\frac{π}{2} x \cdot 2 + 2 πn x \cdot 2 = 1 \cdot 2

Simplificar

\frac{π}{2} x \cdot 2 : π x

\frac{π}{2} x \cdot 2

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 π}{2} x

Eliminar los terminos comunes:

2

= x π

Simplificar

2 πn x \cdot 2 : 4 πn x

2 πn x \cdot 2

Multiplicar los numeros:

2 \cdot 2 = 4

= 4 πn x

Simplificar

1 \cdot 2 : 2

1 \cdot 2

Multiplicar los numeros:

1 \cdot 2 = 2

= 2

π x + 4 πn x = 2

π x + 4 πn x = 2

π x + 4 πn x = 2

Factorizar

π x + 4 πn x : π x (1 + 4 n)

π x + 4 πn x

Factorizar el termino común

x π

= x π (1 + 4 n)

π x (1 + 4 n) = 2

Dividir ambos lados entre

π (1 + 4 n); n \neq = - \frac{1}{4}

π x (1 + 4 n) = 2

Dividir ambos lados entre

π (1 + 4 n); n \neq = - \frac{1}{4}

\frac{π x ( 1 + 4 n )}{π ( 1 + 4 n )} = \frac{2}{π ( 1 + 4 n )}; n \neq = - \frac{1}{4}

Simplificar

x = \frac{2}{π ( 1 + 4 n )}; n \neq = - \frac{1}{4}

x = \frac{2}{π ( 1 + 4 n )}; n \neq = - \frac{1}{4}

Resolver

\frac{1}{x} = \frac{3 π}{2} + 2 πn : x = \frac{2}{π ( 3 + 4 n )}; n \neq = - \frac{3}{4}

\frac{1}{x} = \frac{3 π}{2} + 2 πn

Multiplicar ambos lados por

x

\frac{1}{x} = \frac{3 π}{2} + 2 πn

Multiplicar ambos lados por

x

\frac{1}{x} x = \frac{3 π}{2} x + 2 πn x

Simplificar

1 = \frac{3 π}{2} x + 2 πn x

1 = \frac{3 π}{2} x + 2 πn x

Intercambiar lados

\frac{3 π}{2} x + 2 πn x = 1

Multiplicar ambos lados por

2

\frac{3 π}{2} x + 2 πn x = 1

Multiplicar ambos lados por

2

\frac{3 π}{2} x \cdot 2 + 2 πn x \cdot 2 = 1 \cdot 2

Simplificar

\frac{3 π}{2} x \cdot 2 + 2 πn x \cdot 2 = 1 \cdot 2

Simplificar

\frac{3 π}{2} x \cdot 2 : 3 π x

\frac{3 π}{2} x \cdot 2

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 \cdot 2 π}{2} x

Eliminar los terminos comunes:

2

= x \cdot 3 π

Simplificar

2 πn x \cdot 2 : 4 πn x

2 πn x \cdot 2

Multiplicar los numeros:

2 \cdot 2 = 4

= 4 πn x

Simplificar

1 \cdot 2 : 2

1 \cdot 2

Multiplicar los numeros:

1 \cdot 2 = 2

= 2

3 π x + 4 πn x = 2

3 π x + 4 πn x = 2

3 π x + 4 πn x = 2

Factorizar

3 π x + 4 πn x : π x (3 + 4 n)

3 π x + 4 πn x

Factorizar el termino común

x π

= x π (3 + 4 n)

π x (3 + 4 n) = 2

Dividir ambos lados entre

π (3 + 4 n); n \neq = - \frac{3}{4}

π x (3 + 4 n) = 2

Dividir ambos lados entre

π (3 + 4 n); n \neq = - \frac{3}{4}

\frac{π x ( 3 + 4 n )}{π ( 3 + 4 n )} = \frac{2}{π ( 3 + 4 n )}; n \neq = - \frac{3}{4}

Simplificar

x = \frac{2}{π ( 3 + 4 n )}; n \neq = - \frac{3}{4}

x = \frac{2}{π ( 3 + 4 n )}; n \neq = - \frac{3}{4}

x = \frac{2}{π ( 1 + 4 n )}, x = \frac{2}{π ( 3 + 4 n )}; n \neq = - \frac{1}{4}, n \neq = - \frac{3}{4}

Soluciones para el rango

0 \leq x < 2 π

x = \frac{2}{π}, x = \frac{2}{3 π}, x = \frac{2}{5 π}, x = \frac{2}{7 π}, x = \frac{2}{9 π}, x = \frac{2}{11 π}, x = \frac{2}{13 π}, x = \frac{2}{15 π}, x = \frac{2}{17 π}, x = \frac{2}{19 π}

cos (\frac{1}{x + 1}) = 0, 0 \leq x < 2 π :

Sin solución

cos (\frac{1}{x + 1}) = 0, 0 \leq x < 2 π

Soluciones generales para

cos (\frac{1}{x + 1}) = 0

cos (x)

tabla de valores periódicos con

2 πn

intervalos:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

\frac{1}{x + 1} = \frac{π}{2} + 2 πn, \frac{1}{x + 1} = \frac{3 π}{2} + 2 πn

\frac{1}{x + 1} = \frac{π}{2} + 2 πn, \frac{1}{x + 1} = \frac{3 π}{2} + 2 πn

Resolver

\frac{1}{x + 1} = \frac{π}{2} + 2 πn : x = \frac{2}{π ( 1 + 4 n )} - \frac{4 n}{1 + 4 n} - \frac{1}{1 + 4 n}; n \neq = - \frac{1}{4}

\frac{1}{x + 1} = \frac{π}{2} + 2 πn

Multiplicar por el mínimo común múltiplo

\frac{1}{x + 1} = \frac{π}{2} + 2 πn

Encontrar el mínimo común múltiplo de

x + 1, 2 : 2 (x + 1)

x + 1, 2

Mínimo común múltiplo (MCM)

Calcular una expresión que este compuesta de factores que aparezcan tanto en

x + 1

2

= 2 (x + 1)

Multiplicar por el mínimo común múltiplo=

2 (x + 1)

\frac{1}{x + 1} \cdot 2 (x + 1) = \frac{π}{2} \cdot 2 (x + 1) + 2 πn \cdot 2 (x + 1)

Simplificar

\frac{1}{x + 1} \cdot 2 (x + 1) = \frac{π}{2} \cdot 2 (x + 1) + 2 πn \cdot 2 (x + 1)

Simplificar

\frac{1}{x + 1} \cdot 2 (x + 1) : 2

\frac{1}{x + 1} \cdot 2 (x + 1)

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2 ( x + 1 )}{x + 1}

Eliminar los terminos comunes:

x + 1

= 1 \cdot 2

Multiplicar los numeros:

1 \cdot 2 = 2

= 2

Simplificar

\frac{π}{2} \cdot 2 (x + 1) : π (x + 1)

\frac{π}{2} \cdot 2 (x + 1)

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 π}{2} (x + 1)

Eliminar los terminos comunes:

2

= (x + 1) π

Simplificar

2 πn \cdot 2 (x + 1) : 4 πn (x + 1)

2 πn \cdot 2 (x + 1)

Multiplicar los numeros:

2 \cdot 2 = 4

= 4 πn (x + 1)

2 = π (x + 1) + 4 πn (x + 1)

2 = π (x + 1) + 4 πn (x + 1)

2 = π (x + 1) + 4 πn (x + 1)

Intercambiar lados

π (x + 1) + 4 πn (x + 1) = 2

Expandir

π (x + 1) : π x + π

π (x + 1)

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = π, b = x, c = 1

= π x + π 1

= π x + 1 π

Multiplicar:

1 π = π

= π x + π

π x + π + 4 πn (x + 1) = 2

Expandir

4 πn (x + 1) : 4 πn x + 4 πn

4 πn (x + 1)

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = 4 πn, b = x, c = 1

= 4 πn x + 4 πn \cdot 1

= 4 πn x + 4 \cdot 1 πn

Multiplicar los numeros:

4 \cdot 1 = 4

= 4 πn x + 4 πn

π x + π + 4 πn x + 4 πn = 2

Mover

4 πn

al lado derecho

π x + π + 4 πn x + 4 πn = 2

Restar

4 πn

de ambos lados

π x + π + 4 πn x + 4 πn - 4 πn = 2 - 4 πn

Simplificar

π x + π + 4 πn x = 2 - 4 πn

π x + π + 4 πn x = 2 - 4 πn

Mover

π

al lado derecho

π x + π + 4 πn x = 2 - 4 πn

Restar

π

de ambos lados

π x + π + 4 πn x - π = 2 - 4 πn - π

Simplificar

π x + 4 πn x = 2 - 4 πn - π

π x + 4 πn x = 2 - 4 πn - π

Factorizar

π x + 4 πn x : π x (1 + 4 n)

π x + 4 πn x

Factorizar el termino común

x π

= x π (1 + 4 n)

π x (1 + 4 n) = 2 - 4 πn - π

Dividir ambos lados entre

π (1 + 4 n); n \neq = - \frac{1}{4}

π x (1 + 4 n) = 2 - 4 πn - π

Dividir ambos lados entre

π (1 + 4 n); n \neq = - \frac{1}{4}

\frac{π x ( 1 + 4 n )}{π ( 1 + 4 n )} = \frac{2}{π ( 1 + 4 n )} - \frac{4 πn}{π ( 1 + 4 n )} - \frac{π}{π ( 1 + 4 n )}; n \neq = - \frac{1}{4}

Simplificar

\frac{π x ( 1 + 4 n )}{π ( 1 + 4 n )} = \frac{2}{π ( 1 + 4 n )} - \frac{4 πn}{π ( 1 + 4 n )} - \frac{π}{π ( 1 + 4 n )}

Simplificar

\frac{π x ( 1 + 4 n )}{π ( 1 + 4 n )} : x

\frac{π x ( 1 + 4 n )}{π ( 1 + 4 n )}

Eliminar los terminos comunes:

π

= \frac{x ( 4 n + 1 )}{1 + 4 n}

Eliminar los terminos comunes:

1 + 4 n

= x

Simplificar

\frac{2}{π ( 1 + 4 n )} - \frac{4 πn}{π ( 1 + 4 n )} - \frac{π}{π ( 1 + 4 n )} : \frac{2}{π ( 1 + 4 n )} - \frac{4 n}{1 + 4 n} - \frac{1}{1 + 4 n}

\frac{2}{π ( 1 + 4 n )} - \frac{4 πn}{π ( 1 + 4 n )} - \frac{π}{π ( 1 + 4 n )}

Cancelar

\frac{4 πn}{π ( 1 + 4 n )} : \frac{4 n}{4 n + 1}

\frac{4 πn}{π ( 1 + 4 n )}

Eliminar los terminos comunes:

π

= \frac{4 n}{1 + 4 n}

= \frac{2}{π ( 4 n + 1 )} - \frac{4 n}{4 n + 1} - \frac{π}{π ( 4 n + 1 )}

Cancelar

\frac{π}{π ( 1 + 4 n )} : \frac{1}{4 n + 1}

\frac{π}{π ( 1 + 4 n )}

Eliminar los terminos comunes:

π

= \frac{1}{1 + 4 n}

= \frac{2}{π ( 4 n + 1 )} - \frac{4 n}{4 n + 1} - \frac{1}{4 n + 1}

x = \frac{2}{π ( 1 + 4 n )} - \frac{4 n}{1 + 4 n} - \frac{1}{1 + 4 n}; n \neq = - \frac{1}{4}

x = \frac{2}{π ( 1 + 4 n )} - \frac{4 n}{1 + 4 n} - \frac{1}{1 + 4 n}; n \neq = - \frac{1}{4}

x = \frac{2}{π ( 1 + 4 n )} - \frac{4 n}{1 + 4 n} - \frac{1}{1 + 4 n}; n \neq = - \frac{1}{4}

Resolver

\frac{1}{x + 1} = \frac{3 π}{2} + 2 πn : x = \frac{2}{π ( 3 + 4 n )} - \frac{4 n}{3 + 4 n} - \frac{3}{3 + 4 n}; n \neq = - \frac{3}{4}

\frac{1}{x + 1} = \frac{3 π}{2} + 2 πn

Multiplicar por el mínimo común múltiplo

\frac{1}{x + 1} = \frac{3 π}{2} + 2 πn

Encontrar el mínimo común múltiplo de

x + 1, 2 : 2 (x + 1)

x + 1, 2

Mínimo común múltiplo (MCM)

Calcular una expresión que este compuesta de factores que aparezcan tanto en

x + 1

2

= 2 (x + 1)

Multiplicar por el mínimo común múltiplo=

2 (x + 1)

\frac{1}{x + 1} \cdot 2 (x + 1) = \frac{3 π}{2} \cdot 2 (x + 1) + 2 πn \cdot 2 (x + 1)

Simplificar

\frac{1}{x + 1} \cdot 2 (x + 1) = \frac{3 π}{2} \cdot 2 (x + 1) + 2 πn \cdot 2 (x + 1)

Simplificar

\frac{1}{x + 1} \cdot 2 (x + 1) : 2

\frac{1}{x + 1} \cdot 2 (x + 1)

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2 ( x + 1 )}{x + 1}

Eliminar los terminos comunes:

x + 1

= 1 \cdot 2

Multiplicar los numeros:

1 \cdot 2 = 2

= 2

Simplificar

\frac{3 π}{2} \cdot 2 (x + 1) : 3 π (x + 1)

\frac{3 π}{2} \cdot 2 (x + 1)

Multiplicar fracciones:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 \cdot 2 π}{2} (x + 1)

Eliminar los terminos comunes:

2

= (x + 1) \cdot 3 π

Simplificar

2 πn \cdot 2 (x + 1) : 4 πn (x + 1)

2 πn \cdot 2 (x + 1)

Multiplicar los numeros:

2 \cdot 2 = 4

= 4 πn (x + 1)

2 = 3 π (x + 1) + 4 πn (x + 1)

2 = 3 π (x + 1) + 4 πn (x + 1)

2 = 3 π (x + 1) + 4 πn (x + 1)

Intercambiar lados

3 π (x + 1) + 4 πn (x + 1) = 2

Expandir

3 π (x + 1) : 3 π x + 3 π

3 π (x + 1)

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = 3 π, b = x, c = 1

= 3 π x + 3 π 1

= 3 π x + 3 \cdot 1 π

Multiplicar los numeros:

3 \cdot 1 = 3

= 3 π x + 3 π

3 π x + 3 π + 4 πn (x + 1) = 2

Expandir

4 πn (x + 1) : 4 πn x + 4 πn

4 πn (x + 1)

Poner los parentesis utilizando:

a (b + c) = ab + a c

a = 4 πn, b = x, c = 1

= 4 πn x + 4 πn \cdot 1

= 4 πn x + 4 \cdot 1 πn

Multiplicar los numeros:

4 \cdot 1 = 4

= 4 πn x + 4 πn

3 π x + 3 π + 4 πn x + 4 πn = 2

Mover

4 πn

al lado derecho

3 π x + 3 π + 4 πn x + 4 πn = 2

Restar

4 πn

de ambos lados

3 π x + 3 π + 4 πn x + 4 πn - 4 πn = 2 - 4 πn

Simplificar

3 π x + 3 π + 4 πn x = 2 - 4 πn

3 π x + 3 π + 4 πn x = 2 - 4 πn

Mover

3 π

al lado derecho

3 π x + 3 π + 4 πn x = 2 - 4 πn

Restar

3 π

de ambos lados

3 π x + 3 π + 4 πn x - 3 π = 2 - 4 πn - 3 π

Simplificar

3 π x + 4 πn x = 2 - 4 πn - 3 π

3 π x + 4 πn x = 2 - 4 πn - 3 π

Factorizar

3 π x + 4 πn x : π x (3 + 4 n)

3 π x + 4 πn x

Factorizar el termino común

x π

= x π (3 + 4 n)

π x (3 + 4 n) = 2 - 4 πn - 3 π

Dividir ambos lados entre

π (3 + 4 n); n \neq = - \frac{3}{4}

π x (3 + 4 n) = 2 - 4 πn - 3 π

Dividir ambos lados entre

π (3 + 4 n); n \neq = - \frac{3}{4}

\frac{π x ( 3 + 4 n )}{π ( 3 + 4 n )} = \frac{2}{π ( 3 + 4 n )} - \frac{4 πn}{π ( 3 + 4 n )} - \frac{3 π}{π ( 3 + 4 n )}; n \neq = - \frac{3}{4}

Simplificar

\frac{π x ( 3 + 4 n )}{π ( 3 + 4 n )} = \frac{2}{π ( 3 + 4 n )} - \frac{4 πn}{π ( 3 + 4 n )} - \frac{3 π}{π ( 3 + 4 n )}

Simplificar

\frac{π x ( 3 + 4 n )}{π ( 3 + 4 n )} : x

\frac{π x ( 3 + 4 n )}{π ( 3 + 4 n )}

Eliminar los terminos comunes:

π

= \frac{x ( 4 n + 3 )}{3 + 4 n}

Eliminar los terminos comunes:

3 + 4 n

= x

Simplificar

\frac{2}{π ( 3 + 4 n )} - \frac{4 πn}{π ( 3 + 4 n )} - \frac{3 π}{π ( 3 + 4 n )} : \frac{2}{π ( 3 + 4 n )} - \frac{4 n}{3 + 4 n} - \frac{3}{3 + 4 n}

\frac{2}{π ( 3 + 4 n )} - \frac{4 πn}{π ( 3 + 4 n )} - \frac{3 π}{π ( 3 + 4 n )}

Cancelar

\frac{4 πn}{π ( 3 + 4 n )} : \frac{4 n}{4 n + 3}

\frac{4 πn}{π ( 3 + 4 n )}

Eliminar los terminos comunes:

π

= \frac{4 n}{3 + 4 n}

= \frac{2}{π ( 4 n + 3 )} - \frac{4 n}{4 n + 3} - \frac{3 π}{π ( 4 n + 3 )}

Cancelar

\frac{3 π}{π ( 3 + 4 n )} : \frac{3}{4 n + 3}

\frac{3 π}{π ( 3 + 4 n )}

Eliminar los terminos comunes:

π

= \frac{3}{3 + 4 n}

= \frac{2}{π ( 4 n + 3 )} - \frac{4 n}{4 n + 3} - \frac{3}{4 n + 3}

x = \frac{2}{π ( 3 + 4 n )} - \frac{4 n}{3 + 4 n} - \frac{3}{3 + 4 n}; n \neq = - \frac{3}{4}

x = \frac{2}{π ( 3 + 4 n )} - \frac{4 n}{3 + 4 n} - \frac{3}{3 + 4 n}; n \neq = - \frac{3}{4}

x = \frac{2}{π ( 3 + 4 n )} - \frac{4 n}{3 + 4 n} - \frac{3}{3 + 4 n}; n \neq = - \frac{3}{4}

x = \frac{2}{π ( 1 + 4 n )} - \frac{4 n}{1 + 4 n} - \frac{1}{1 + 4 n}, x = \frac{2}{π ( 3 + 4 n )} - \frac{4 n}{3 + 4 n} - \frac{3}{3 + 4 n}; n \neq = - \frac{1}{4}, n \neq = - \frac{3}{4}

Soluciones para el rango

0 \leq x < 2 π

Sin soluci \overset{o}{ˊ} n

Combinar toda las soluciones

x = \frac{2}{π}, x = \frac{2}{3 π}, x = \frac{2}{5 π}, x = \frac{2}{7 π}, x = \frac{2}{9 π}, x = \frac{2}{11 π}, x = \frac{2}{13 π}, x = \frac{2}{15 π}, x = \frac{2}{17 π}, x = \frac{2}{19 π}

\frac{2}{19 π}, \frac{- 9 π + 81 π ^{2} + 36 π}{18 π}, \frac{2}{17 π}, \frac{- 2 π + 4 π ^{2} + 2 π}{4 π}, \frac{2}{15 π}, \frac{- 7 π + 49 π ^{2} + 28 π}{14 π}, \frac{2}{13 π}, \frac{- 3 π + 9 π ^{2} + 6 π}{6 π}, \frac{2}{11 π}, \frac{- 5 π + 25 π ^{2} + 20 π}{10 π}, \frac{2}{9 π}, \frac{- 2 π + 4 π ^{2} + 4 π}{4 π}, \frac{2}{7 π}, \frac{- 3 π + 9 π ^{2} + 12 π}{6 π}, \frac{2}{5 π}, \frac{- π + π ^{2} + 2 π}{2 π}, \frac{2}{3 π}, \frac{π ^{2} + 4 π - π}{2 π}, \frac{2}{π}

Identificar los intervalos

0 < x < \frac{2}{19 π}, \frac{2}{19 π} < x < \frac{- 9 π + 81 π ^{2} + 36 π}{18 π}, \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} < x < \frac{2}{17 π}, \frac{2}{17 π} < x < \frac{- 2 π + 4 π ^{2} + 2 π}{4 π}, \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} < x < \frac{2}{15 π}, \frac{2}{15 π} < x < \frac{- 7 π + 49 π ^{2} + 28 π}{14 π}, \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} < x < \frac{2}{13 π}, \frac{2}{13 π} < x < \frac{- 3 π + 9 π ^{2} + 6 π}{6 π}, \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} < x < \frac{2}{11 π}, \frac{2}{11 π} < x < \frac{- 5 π + 25 π ^{2} + 20 π}{10 π}, \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} < x < \frac{2}{9 π}, \frac{2}{9 π} < x < \frac{- 2 π + 4 π ^{2} + 4 π}{4 π}, \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} < x < \frac{2}{7 π}, \frac{2}{7 π} < x < \frac{- 3 π + 9 π ^{2} + 12 π}{6 π}, \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} < x < \frac{2}{5 π}, \frac{2}{5 π} < x < \frac{- π + π ^{2} + 2 π}{2 π}, \frac{- π + π ^{2} + 2 π}{2 π} < x < \frac{2}{3 π}, \frac{2}{3 π} < x < \frac{π ^{2} + 4 π - π}{2 π}, \frac{π ^{2} + 4 π - π}{2 π} < x < \frac{2}{π}, \frac{2}{π} < x < 2 π

Resumir en una tabla:

Identificar los intervalos que cumplen la condición:

\leq 0

0 < x < \frac{2}{19 π} or x = \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} < x < \frac{2}{17 π} or x = \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} < x < \frac{2}{15 π} or x = \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} < x < \frac{2}{13 π} or x = \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} < x < \frac{2}{11 π} or x = \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} < x < \frac{2}{9 π} or x = \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} < x < \frac{2}{7 π} or x = \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} < x < \frac{2}{5 π} or x = \frac{- π + π ^{2} + 2 π}{2 π} or \frac{- π + π ^{2} + 2 π}{2 π} < x < \frac{2}{3 π} or x = \frac{π ^{2} + 4 π - π}{2 π} or \frac{π ^{2} + 4 π - π}{2 π} < x < \frac{2}{π}

Mezclar intervalos sobrepuestos

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} \leq x < \frac{2}{5 π} or \frac{- π + π ^{2} + 2 π}{2 π} \leq x < \frac{2}{3 π} or x = \frac{π ^{2} + 4 π - π}{2 π} or \frac{π ^{2} + 4 π - π}{2 π} < x < \frac{2}{π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π}

x = \frac{- 9 π + 81 π ^{2} + 36 π}{18 π}

0 < x < \frac{2}{19 π} or x = \frac{- 9 π + 81 π ^{2} + 36 π}{18 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or x = \frac{- 9 π + 81 π ^{2} + 36 π}{18 π}

\frac{- 9 π + 81 π ^{2} + 36 π}{18 π} < x < \frac{2}{17 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π}

x = \frac{- 2 π + 4 π ^{2} + 2 π}{4 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or x = \frac{- 2 π + 4 π ^{2} + 2 π}{4 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or x = \frac{- 2 π + 4 π ^{2} + 2 π}{4 π}

\frac{- 2 π + 4 π ^{2} + 2 π}{4 π} < x < \frac{2}{15 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π}

x = \frac{- 7 π + 49 π ^{2} + 28 π}{14 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or x = \frac{- 7 π + 49 π ^{2} + 28 π}{14 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or x = \frac{- 7 π + 49 π ^{2} + 28 π}{14 π}

\frac{- 7 π + 49 π ^{2} + 28 π}{14 π} < x < \frac{2}{13 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π}

x = \frac{- 3 π + 9 π ^{2} + 6 π}{6 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or x = \frac{- 3 π + 9 π ^{2} + 6 π}{6 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or x = \frac{- 3 π + 9 π ^{2} + 6 π}{6 π}

\frac{- 3 π + 9 π ^{2} + 6 π}{6 π} < x < \frac{2}{11 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π}

x = \frac{- 5 π + 25 π ^{2} + 20 π}{10 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or x = \frac{- 5 π + 25 π ^{2} + 20 π}{10 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or x = \frac{- 5 π + 25 π ^{2} + 20 π}{10 π}

\frac{- 5 π + 25 π ^{2} + 20 π}{10 π} < x < \frac{2}{9 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π}

x = \frac{- 2 π + 4 π ^{2} + 4 π}{4 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or x = \frac{- 2 π + 4 π ^{2} + 4 π}{4 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or x = \frac{- 2 π + 4 π ^{2} + 4 π}{4 π}

\frac{- 2 π + 4 π ^{2} + 4 π}{4 π} < x < \frac{2}{7 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π}

x = \frac{- 3 π + 9 π ^{2} + 12 π}{6 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or x = \frac{- 3 π + 9 π ^{2} + 12 π}{6 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or x = \frac{- 3 π + 9 π ^{2} + 12 π}{6 π}

\frac{- 3 π + 9 π ^{2} + 12 π}{6 π} < x < \frac{2}{5 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} \leq x < \frac{2}{5 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} \leq x < \frac{2}{5 π}

x = \frac{- π + π ^{2} + 2 π}{2 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} \leq x < \frac{2}{5 π} or x = \frac{- π + π ^{2} + 2 π}{2 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} \leq x < \frac{2}{5 π} or x = \frac{- π + π ^{2} + 2 π}{2 π}

\frac{- π + π ^{2} + 2 π}{2 π} < x < \frac{2}{3 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} \leq x < \frac{2}{5 π} or \frac{- π + π ^{2} + 2 π}{2 π} \leq x < \frac{2}{3 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} \leq x < \frac{2}{5 π} or \frac{- π + π ^{2} + 2 π}{2 π} \leq x < \frac{2}{3 π}

x = \frac{π ^{2} + 4 π - π}{2 π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} \leq x < \frac{2}{5 π} or \frac{- π + π ^{2} + 2 π}{2 π} \leq x < \frac{2}{3 π} or x = \frac{π ^{2} + 4 π - π}{2 π}

La unión de dos intervalos comprende a los conjuntos numéricos que están en el primero y en el segundo

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} \leq x < \frac{2}{5 π} or \frac{- π + π ^{2} + 2 π}{2 π} \leq x < \frac{2}{3 π} or x = \frac{π ^{2} + 4 π - π}{2 π}

\frac{π ^{2} + 4 π - π}{2 π} < x < \frac{2}{π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} \leq x < \frac{2}{5 π} or \frac{- π + π ^{2} + 2 π}{2 π} \leq x < \frac{2}{3 π} or \frac{π ^{2} + 4 π - π}{2 π} \leq x < \frac{2}{π}

0 < x < \frac{2}{19 π} or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} \leq x < \frac{2}{17 π} or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} \leq x < \frac{2}{15 π} or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} \leq x < \frac{2}{13 π} or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} \leq x < \frac{2}{11 π} or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} \leq x < \frac{2}{9 π} or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} \leq x < \frac{2}{7 π} or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} \leq x < \frac{2}{5 π} or \frac{- π + π ^{2} + 2 π}{2 π} \leq x < \frac{2}{3 π} or \frac{π ^{2} + 4 π - π}{2 π} \leq x < \frac{2}{π}

Utilizar la periodicidad de

tan (\frac{1}{x}) - tan (\frac{1}{x + 1})

2 πn < x < \frac{2}{19 π} + 2 πn or \frac{- 9 π + 81 π ^{2} + 36 π}{18 π} + 2 πn \leq x < \frac{2}{17 π} + 2 πn or \frac{- 2 π + 4 π ^{2} + 2 π}{4 π} + 2 πn \leq x < \frac{2}{15 π} + 2 πn or \frac{- 7 π + 49 π ^{2} + 28 π}{14 π} + 2 πn \leq x < \frac{2}{13 π} + 2 πn or \frac{- 3 π + 9 π ^{2} + 6 π}{6 π} + 2 πn \leq x < \frac{2}{11 π} + 2 πn or \frac{- 5 π + 25 π ^{2} + 20 π}{10 π} + 2 πn \leq x < \frac{2}{9 π} + 2 πn or \frac{- 2 π + 4 π ^{2} + 4 π}{4 π} + 2 πn \leq x < \frac{2}{7 π} + 2 πn or \frac{- 3 π + 9 π ^{2} + 12 π}{6 π} + 2 πn \leq x < \frac{2}{5 π} + 2 πn or \frac{- π + π ^{2} + 2 π}{2 π} + 2 πn \leq x < \frac{2}{3 π} + 2 πn or \frac{π ^{2} + 4 π - π}{2 π} + 2 πn \leq x < \frac{2}{π} + 2 πn

Ejemplos populares

cos(x^4)+sin(x^4)>= 0.5

cos (x^{4}) + sin (x^{4}) \geq 0.5

1-tan(x)<2

1 - tan (x) < 2

2sin(x)-1>=-3

2 sin (x) - 1 \geq - 3

30<= sqrt(980)cos(θ)3.19

30 \leq 980 cos (θ) 3.19

2sin(x)>sqrt(3)

2 sin (x) > 3