English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3tan(x)+cot(x)<5sin(x)

Solución

3 tan (x) + cot (x) < 5 sin (x)

Solución

\frac{π}{2} + 2 πn < x < π + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

Notación de intervalos

(\frac{π}{2} + 2 πn, π + 2 πn) \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn)

Notación decimal

1.57079 \dots + 2 πn < x < 3.14159 \dots + 2 πn or 4.71238 \dots + 2 πn < x < 6.28318 \dots + 2 πn

Pasos de solución

3 tan (x) + cot (x) < 5 sin (x)

Mover

5 sin (x)

al lado izquierdo

3 tan (x) + cot (x) < 5 sin (x)

Restar

5 sin (x)

de ambos lados

3 tan (x) + cot (x) - 5 sin (x) < 5 sin (x) - 5 sin (x)

3 tan (x) + cot (x) - 5 sin (x) < 0

3 tan (x) + cot (x) - 5 sin (x) < 0

Periodicidad de

3 tan (x) + cot (x) - 5 sin (x) : 2 π

La periodicidad combinada de la suma de funciones periódicas es el mínimo común múltiplo de los períodos

3 tan (x), cot (x), 5 sin (x)

Periodicidad de

3 tan (x) : π

La periodicidad de

a \cdot : t an (b x + c) + d = \frac{periodicidad de tan ( x )}{∣ b ∣}

La periodicidad de

: t an (x)

π

= \frac{π}{∣ 1 ∣}

Simplificar

= π

Periodicidad de

cot (x) : π

La periodicidad de

: co t (x)

π

= π

Periodicidad de

5 sin (x) : 2 π

La periodicidad de

a \cdot : s in (b x + c) + d = \frac{periodicidad de sin ( x )}{∣ b ∣}

La periodicidad de

: s in (x)

2 π

= \frac{2 π}{∣ 1 ∣}

Simplificar

= 2 π

Combinar períodos:

π, π, 2 π

= 2 π

Expresar con seno, coseno

3 tan (x) + cot (x) - 5 sin (x) < 0

Utilizar la identidad trigonométrica básica:

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

3 \cdot \frac{sin ( x )}{cos ( x )} + cot (x) - 5 sin (x) < 0

Utilizar la identidad trigonométrica básica:

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

3 \cdot \frac{sin ( x )}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - 5 sin (x) < 0

3 \cdot \frac{sin ( x )}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - 5 sin (x) < 0

Simplificar

3 \cdot \frac{sin ( x )}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - 5 sin (x) : \frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )}

3 \cdot \frac{sin ( x )}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - 5 sin (x)

Multiplicar

3 \cdot \frac{sin ( x )}{cos ( x )} : \frac{3 sin ( x )}{cos ( x )}

3 \cdot \frac{sin ( x )}{cos ( x )}

Multiplicar fracciones:

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

= \frac{sin ( x ) \cdot 3}{cos ( x )}

= \frac{3 sin ( x )}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - 5 sin (x)

Convertir a fracción:

5 sin (x) = \frac{5 sin ( x )}{1}

= \frac{sin ( x ) \cdot 3}{cos ( x )} + \frac{cos ( x )}{sin ( x )} - \frac{5 sin ( x )}{1}

Mínimo común múltiplo de

cos (x), sin (x), 1 : cos (x) sin (x)

cos (x), sin (x), 1

Mínimo común múltiplo (MCM)

Calcular una expresión que este compuesta de factores que aparezcan en al menos una de las expresiones factorizadas

= cos (x) sin (x)

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 ( x ) \cdot 3}{cos ( x )} :

multiplicar el denominador y el numerador por

sin (x)

\frac{sin ( x ) \cdot 3}{cos ( x )} = \frac{sin ( x ) \cdot 3 sin ( x )}{cos ( x ) sin ( x )} = \frac{3 sin ^{2} ( x )}{cos ( x ) sin ( x )}

Para

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

multiplicar el denominador y el numerador por

cos (x)

\frac{cos ( x )}{sin ( x )} = \frac{cos ( x ) cos ( x )}{sin ( x ) cos ( x )} = \frac{cos ^{2} ( x )}{cos ( x ) sin ( x )}

Para

\frac{5 sin ( x )}{1} :

multiplicar el denominador y el numerador por

cos (x) sin (x)

\frac{5 sin ( x )}{1} = \frac{5 sin ( x ) cos ( x ) sin ( x )}{1 \cdot cos ( x ) sin ( x )} = \frac{5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )}

= \frac{3 sin ^{2} ( x )}{cos ( x ) sin ( x )} + \frac{cos ^{2} ( x )}{cos ( x ) sin ( x )} - \frac{5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )}

\frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )} < 0

Encontrar los ceros y puntos indefinidos de

\frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )}

para

0 \leq x < 2 π

Para encontrar los ceros, transformar la desigualdad a 0

\frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )} = 0

\frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )} = 0, 0 \leq x < 2 π :

Sin solución para

x \in R

\frac{3 sin ^{2} ( x ) + cos ^{2} ( x ) - 5 sin ^{2} ( x ) cos ( x )}{cos ( x ) sin ( x )} = 0, 0 \leq x < 2 π

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

3 sin^{2} (x) + cos^{2} (x) - 5 sin^{2} (x) cos (x) = 0

Re-escribir usando identidades trigonométricas

cos^{2} (x) + 3 sin^{2} (x) - 5 cos (x) sin^{2} (x)

Utilizar la identidad pitagórica:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= cos^{2} (x) + 3 (1 - cos^{2} (x)) - 5 cos (x) (1 - cos^{2} (x))

Simplificar

cos^{2} (x) + 3 (1 - cos^{2} (x)) - 5 cos (x) (1 - cos^{2} (x)) : - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

cos^{2} (x) + 3 (1 - cos^{2} (x)) - 5 cos (x) (1 - cos^{2} (x))

Expandir

3 (1 - cos^{2} (x)) : 3 - 3 cos^{2} (x)

3 (1 - cos^{2} (x))

Poner los parentesis utilizando:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (x)

= 3 \cdot 1 - 3 cos^{2} (x)

Multiplicar los numeros:

3 \cdot 1 = 3

= 3 - 3 cos^{2} (x)

= cos^{2} (x) + 3 - 3 cos^{2} (x) - 5 cos (x) (1 - cos^{2} (x))

Expandir

- 5 cos (x) (1 - cos^{2} (x)) : - 5 cos (x) + 5 cos^{3} (x)

- 5 cos (x) (1 - cos^{2} (x))

Poner los parentesis utilizando:

a (b - c) = ab - a c

a = - 5 cos (x), b = 1, c = cos^{2} (x)

= - 5 cos (x) \cdot 1 - (- 5 cos (x)) cos^{2} (x)

Aplicar las reglas de los signos

- (- a) = a

= - 5 \cdot 1 \cdot cos (x) + 5 cos^{2} (x) cos (x)

Simplificar

- 5 \cdot 1 \cdot cos (x) + 5 cos^{2} (x) cos (x) : - 5 cos (x) + 5 cos^{3} (x)

- 5 \cdot 1 \cdot cos (x) + 5 cos^{2} (x) cos (x)

5 \cdot 1 \cdot cos (x) = 5 cos (x)

5 \cdot 1 \cdot cos (x)

Multiplicar los numeros:

5 \cdot 1 = 5

= 5 cos (x)

5 cos^{2} (x) cos (x) = 5 cos^{3} (x)

5 cos^{2} (x) cos (x)

Aplicar las leyes de los exponentes:

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

cos^{2} (x) cos (x) = cos^{2 + 1} (x)

= 5 cos^{2 + 1} (x)

Sumar:

2 + 1 = 3

= 5 cos^{3} (x)

= - 5 cos (x) + 5 cos^{3} (x)

= - 5 cos (x) + 5 cos^{3} (x)

= cos^{2} (x) + 3 - 3 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x)

Simplificar

cos^{2} (x) + 3 - 3 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) : - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

cos^{2} (x) + 3 - 3 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x)

Agrupar términos semejantes

= cos^{2} (x) - 3 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

Sumar elementos similares:

cos^{2} (x) - 3 cos^{2} (x) = - 2 cos^{2} (x)

= - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

= - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

= - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) + 3

3 - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) = 0

Usando el método de sustitución

3 - 2 cos^{2} (x) - 5 cos (x) + 5 cos^{3} (x) = 0

Sea:

cos (x) = u

3 - 2 u^{2} - 5 u + 5 u^{3} = 0

3 - 2 u^{2} - 5 u + 5 u^{3} = 0 : u \approx - 1.06603 \dots

3 - 2 u^{2} - 5 u + 5 u^{3} = 0

Escribir en la forma binómica

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

5 u^{3} - 2 u^{2} - 5 u + 3 = 0

Encontrar una solución para

5 u^{3} - 2 u^{2} - 5 u + 3 = 0

utilizando el método de Newton-Raphson:

u \approx - 1.06603 \dots

5 u^{3} - 2 u^{2} - 5 u + 3 = 0

Definición del método de Newton-Raphson

f (u) = 5 u^{3} - 2 u^{2} - 5 u + 3

Hallar

f^{^{'}} (u) : 15 u^{2} - 4 u - 5

\frac{d}{d u} (5 u^{3} - 2 u^{2} - 5 u + 3)

Aplicar la regla de la suma/diferencia:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (5 u^{3}) - \frac{d}{d u} (2 u^{2}) - \frac{d}{d u} (5 u) + \frac{d}{d u} (3)

\frac{d}{d u} (5 u^{3}) = 15 u^{2}

\frac{d}{d u} (5 u^{3})

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d}{d u} (u^{3})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 5 \cdot 3 u^{3 - 1}

Simplificar

= 15 u^{2}

\frac{d}{d u} (2 u^{2}) = 4 u

\frac{d}{d u} (2 u^{2})

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{2})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 2 u^{2 - 1}

Simplificar

= 4 u

\frac{d}{d u} (5 u) = 5

\frac{d}{d u} (5 u)

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d u}{d u}

Aplicar la regla de derivación:

\frac{d u}{d u} = 1

= 5 \cdot 1

Simplificar

= 5

\frac{d}{d u} (3) = 0

\frac{d}{d u} (3)

Derivada de una constante:

\frac{d}{d x} (a) = 0

= 0

= 15 u^{2} - 4 u - 5 + 0

Simplificar

= 15 u^{2} - 4 u - 5

Sea

u_{0} = - 1

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = - 1.07142 \dots : Δ u_{1} = 0.07142 \dots

f (u_{0}) = 5 (- 1)^{3} - 2 (- 1)^{2} - 5 (- 1) + 3 = 1

f^{^{'}} (u_{0}) = 15 (- 1)^{2} - 4 (- 1) - 5 = 14

u_{1} = - 1.07142 \dots

Δ u_{1} = ∣ - 1.07142 \dots - (- 1) ∣ = 0.07142 \dots

Δ u_{1} = 0.07142 \dots

u_{2} = - 1.06606 \dots : Δ u_{2} = 0.00536 \dots

f (u_{1}) = 5 (- 1.07142 \dots)^{3} - 2 (- 1.07142 \dots)^{2} - 5 (- 1.07142 \dots) + 3 = - 0.08855 \dots

f^{^{'}} (u_{1}) = 15 (- 1.07142 \dots)^{2} - 4 (- 1.07142 \dots) - 5 = 16.50510 \dots

u_{2} = - 1.06606 \dots

Δ u_{2} = ∣ - 1.06606 \dots - (- 1.07142 \dots) ∣ = 0.00536 \dots

Δ u_{2} = 0.00536 \dots

u_{3} = - 1.06603 \dots : Δ u_{3} = 0.00003 \dots

f (u_{2}) = 5 (- 1.06606 \dots)^{3} - 2 (- 1.06606 \dots)^{2} - 5 (- 1.06606 \dots) + 3 = - 0.00051 \dots

f^{^{'}} (u_{2}) = 15 (- 1.06606 \dots)^{2} - 4 (- 1.06606 \dots) - 5 = 16.31161 \dots

u_{3} = - 1.06603 \dots

Δ u_{3} = ∣ - 1.06603 \dots - (- 1.06606 \dots) ∣ = 0.00003 \dots

Δ u_{3} = 0.00003 \dots

u_{4} = - 1.06603 \dots : Δ u_{4} = 1.11867 E - 9

f (u_{3}) = 5 (- 1.06603 \dots)^{3} - 2 (- 1.06603 \dots)^{2} - 5 (- 1.06603 \dots) + 3 = - 1.8246 E - 8

f^{^{'}} (u_{3}) = 15 (- 1.06603 \dots)^{2} - 4 (- 1.06603 \dots) - 5 = 16.31046 \dots

u_{4} = - 1.06603 \dots

Δ u_{4} = ∣ - 1.06603 \dots - (- 1.06603 \dots) ∣ = 1.11867 E - 9

Δ u_{4} = 1.11867 E - 9

u \approx - 1.06603 \dots

Aplicar la división larga

Eq u a t i o n 0 : \frac{5 u ^{3} - 2 u ^{2} - 5 u + 3}{u + 1.06603 \dots} = 5 u^{2} - 7.33015 \dots u + 2.81417 \dots

5 u^{2} - 7.33015 \dots u + 2.81417 \dots \approx 0

Encontrar una solución para

5 u^{2} - 7.33015 \dots u + 2.81417 \dots = 0

utilizando el método de Newton-Raphson:

Sin solución para

u \in R

5 u^{2} - 7.33015 \dots u + 2.81417 \dots = 0

Definición del método de Newton-Raphson

f (u) = 5 u^{2} - 7.33015 \dots u + 2.81417 \dots

Hallar

f^{^{'}} (u) : 10 u - 7.33015 \dots

\frac{d}{d u} (5 u^{2} - 7.33015 \dots u + 2.81417 \dots)

Aplicar la regla de la suma/diferencia:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (5 u^{2}) - \frac{d}{d u} (7.33015 \dots u) + \frac{d}{d u} (2.81417 \dots)

\frac{d}{d u} (5 u^{2}) = 10 u

\frac{d}{d u} (5 u^{2})

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 5 \frac{d}{d u} (u^{2})

Aplicar la regla de la potencia:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 5 \cdot 2 u^{2 - 1}

Simplificar

= 10 u

\frac{d}{d u} (7.33015 \dots u) = 7.33015 \dots

\frac{d}{d u} (7.33015 \dots u)

Sacar la constante:

(a \cdot f)^{'} = a \cdot f^{'}

= 7.33015 \dots \frac{d u}{d u}

Aplicar la regla de derivación:

\frac{d u}{d u} = 1

= 7.33015 \dots \cdot 1

Simplificar

= 7.33015 \dots

\frac{d}{d u} (2.81417 \dots) = 0

\frac{d}{d u} (2.81417 \dots)

Derivada de una constante:

\frac{d}{d x} (a) = 0

= 0

= 10 u - 7.33015 \dots + 0

Simplificar

= 10 u - 7.33015 \dots

Sea

u_{0} = 0

Calcular

u_{n + 1}

hasta que

Δ u_{n + 1} < 0.000001

u_{1} = 0.38391 \dots : Δ u_{1} = 0.38391 \dots

f (u_{0}) = 5 \cdot 0^{2} - 7.33015 \dots \cdot 0 + 2.81417 \dots = 2.81417 \dots

f^{^{'}} (u_{0}) = 10 \cdot 0 - 7.33015 \dots = - 7.33015 \dots

u_{1} = 0.38391 \dots

Δ u_{1} = ∣ 0.38391 \dots - 0 ∣ = 0.38391 \dots

Δ u_{1} = 0.38391 \dots

u_{2} = 0.59502 \dots : Δ u_{2} = 0.21110 \dots

f (u_{1}) = 5 \cdot 0.38391 \dots^{2} - 7.33015 \dots \cdot 0.38391 \dots + 2.81417 \dots = 0.73696 \dots

f^{^{'}} (u_{1}) = 10 \cdot 0.38391 \dots - 7.33015 \dots = - 3.49098 \dots

u_{2} = 0.59502 \dots

Δ u_{2} = ∣ 0.59502 \dots - 0.38391 \dots ∣ = 0.21110 \dots

Δ u_{2} = 0.21110 \dots

u_{3} = 0.75649 \dots : Δ u_{3} = 0.16147 \dots

f (u_{2}) = 5 \cdot 0.59502 \dots^{2} - 7.33015 \dots \cdot 0.59502 \dots + 2.81417 \dots = 0.22282 \dots

f^{^{'}} (u_{2}) = 10 \cdot 0.59502 \dots - 7.33015 \dots = - 1.37992 \dots

u_{3} = 0.75649 \dots

Δ u_{3} = ∣ 0.75649 \dots - 0.59502 \dots ∣ = 0.16147 \dots

Δ u_{3} = 0.16147 \dots

u_{4} = 0.20133 \dots : Δ u_{4} = 0.55516 \dots

f (u_{3}) = 5 \cdot 0.75649 \dots^{2} - 7.33015 \dots \cdot 0.75649 \dots + 2.81417 \dots = 0.13037 \dots

f^{^{'}} (u_{3}) = 10 \cdot 0.75649 \dots - 7.33015 \dots = 0.23484 \dots

u_{4} = 0.20133 \dots

Δ u_{4} = ∣ 0.20133 \dots - 0.75649 \dots ∣ = 0.55516 \dots

Δ u_{4} = 0.55516 \dots

u_{5} = 0.49118 \dots : Δ u_{5} = 0.28984 \dots

f (u_{4}) = 5 \cdot 0.20133 \dots^{2} - 7.33015 \dots \cdot 0.20133 \dots + 2.81417 \dots = 1.54101 \dots

f^{^{'}} (u_{4}) = 10 \cdot 0.20133 \dots - 7.33015 \dots = - 5.31676 \dots

u_{5} = 0.49118 \dots

Δ u_{5} = ∣ 0.49118 \dots - 0.20133 \dots ∣ = 0.28984 \dots

Δ u_{5} = 0.28984 \dots

u_{6} = 0.66486 \dots : Δ u_{6} = 0.17368 \dots

f (u_{5}) = 5 \cdot 0.49118 \dots^{2} - 7.33015 \dots \cdot 0.49118 \dots + 2.81417 \dots = 0.42003 \dots

f^{^{'}} (u_{5}) = 10 \cdot 0.49118 \dots - 7.33015 \dots = - 2.41835 \dots

u_{6} = 0.66486 \dots

Δ u_{6} = ∣ 0.66486 \dots - 0.49118 \dots ∣ = 0.17368 \dots

Δ u_{6} = 0.17368 \dots

u_{7} = 0.88620 \dots : Δ u_{7} = 0.22133 \dots

f (u_{6}) = 5 \cdot 0.66486 \dots^{2} - 7.33015 \dots \cdot 0.66486 \dots + 2.81417 \dots = 0.15083 \dots

f^{^{'}} (u_{6}) = 10 \cdot 0.66486 \dots - 7.33015 \dots = - 0.68147 \dots

u_{7} = 0.88620 \dots

Δ u_{7} = ∣ 0.88620 \dots - 0.66486 \dots ∣ = 0.22133 \dots

Δ u_{7} = 0.22133 \dots

u_{8} = 0.72630 \dots : Δ u_{8} = 0.15990 \dots

f (u_{7}) = 5 \cdot 0.88620 \dots^{2} - 7.33015 \dots \cdot 0.88620 \dots + 2.81417 \dots = 0.24495 \dots

f^{^{'}} (u_{7}) = 10 \cdot 0.88620 \dots - 7.33015 \dots = 1.53190 \dots

u_{8} = 0.72630 \dots

Δ u_{8} = ∣ 0.72630 \dots - 0.88620 \dots ∣ = 0.15990 \dots

Δ u_{8} = 0.15990 \dots

u_{9} = 2.63145 \dots : Δ u_{9} = 1.90514 \dots

f (u_{8}) = 5 \cdot 0.72630 \dots^{2} - 7.33015 \dots \cdot 0.72630 \dots + 2.81417 \dots = 0.12784 \dots

f^{^{'}} (u_{8}) = 10 \cdot 0.72630 \dots - 7.33015 \dots = - 0.06710 \dots

u_{9} = 2.63145 \dots

Δ u_{9} = ∣ 2.63145 \dots - 0.72630 \dots ∣ = 1.90514 \dots

Δ u_{9} = 1.90514 \dots

u_{10} = 1.67551 \dots : Δ u_{10} = 0.95594 \dots

f (u_{9}) = 5 \cdot 2.63145 \dots^{2} - 7.33015 \dots \cdot 2.63145 \dots + 2.81417 \dots = 18.14798 \dots

f^{^{'}} (u_{9}) = 10 \cdot 2.63145 \dots - 7.33015 \dots = 18.98439 \dots

u_{10} = 1.67551 \dots

Δ u_{10} = ∣ 1.67551 \dots - 2.63145 \dots ∣ = 0.95594 \dots

Δ u_{10} = 0.95594 \dots

u_{11} = 1.19072 \dots : Δ u_{11} = 0.48478 \dots

f (u_{10}) = 5 \cdot 1.67551 \dots^{2} - 7.33015 \dots \cdot 1.67551 \dots + 2.81417 \dots = 4.56912 \dots

f^{^{'}} (u_{10}) = 10 \cdot 1.67551 \dots - 7.33015 \dots = 9.42497 \dots

u_{11} = 1.19072 \dots

Δ u_{11} = ∣ 1.19072 \dots - 1.67551 \dots ∣ = 0.48478 \dots

Δ u_{11} = 0.48478 \dots

u_{12} = 0.93398 \dots : Δ u_{12} = 0.25673 \dots

f (u_{11}) = 5 \cdot 1.19072 \dots^{2} - 7.33015 \dots \cdot 1.19072 \dots + 2.81417 \dots = 1.17510 \dots

f^{^{'}} (u_{11}) = 10 \cdot 1.19072 \dots - 7.33015 \dots = 4.57708 \dots

u_{12} = 0.93398 \dots

Δ u_{12} = ∣ 0.93398 \dots - 1.19072 \dots ∣ = 0.25673 \dots

Δ u_{12} = 0.25673 \dots

u_{13} = 0.77000 \dots : Δ u_{13} = 0.16398 \dots

f (u_{12}) = 5 \cdot 0.93398 \dots^{2} - 7.33015 \dots \cdot 0.93398 \dots + 2.81417 \dots = 0.32956 \dots

f^{^{'}} (u_{12}) = 10 \cdot 0.93398 \dots - 7.33015 \dots = 2.00972 \dots

u_{13} = 0.77000 \dots

Δ u_{13} = ∣ 0.77000 \dots - 0.93398 \dots ∣ = 0.16398 \dots

Δ u_{13} = 0.16398 \dots

u_{14} = 0.40647 \dots : Δ u_{14} = 0.36352 \dots

f (u_{13}) = 5 \cdot 0.77000 \dots^{2} - 7.33015 \dots \cdot 0.77000 \dots + 2.81417 \dots = 0.13445 \dots

f^{^{'}} (u_{13}) = 10 \cdot 0.77000 \dots - 7.33015 \dots = 0.36986 \dots

u_{14} = 0.40647 \dots

Δ u_{14} = ∣ 0.40647 \dots - 0.77000 \dots ∣ = 0.36352 \dots

Δ u_{14} = 0.36352 \dots

No se puede encontrar solución

La solución es

u \approx - 1.06603 \dots

Sustituir en la ecuación

u = cos (x)

cos (x) \approx - 1.06603 \dots

cos (x) \approx - 1.06603 \dots

cos (x) = - 1.06603 \dots, 0 \leq x < 2 π :

Sin solución

cos (x) = - 1.06603 \dots, 0 \leq x < 2 π

- 1 \leq cos (x) \leq 1

Sin soluci \overset{o}{ˊ} n

Combinar toda las soluciones

Sin soluci \overset{o}{ˊ} n para x \in R

Encontrar los puntos indefinidos:

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π

Encontrar los ceros del denominador

cos (x) sin (x) = 0

Resolver cada parte por separado

cos (x) = 0 or sin (x) = 0

cos (x) = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x < 2 π

Soluciones generales para

cos (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}

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

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

Soluciones para el rango

0 \leq x < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

Soluciones generales para

sin (x) = 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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Resolver

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Soluciones para el rango

0 \leq x < 2 π

x = 0, x = π

Combinar toda las soluciones

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π

0, \frac{π}{2}, π, \frac{3 π}{2}

Identificar los intervalos

0 < x < \frac{π}{2}, \frac{π}{2} < x < π, π < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

Resumir en una tabla:

3 sin^{2} (x) + cos^{2} (x) - 5 sin^{2} (x) cos (x) cos (x) sin (x) \frac{3 s i n ^{2} ( x ) + c o s ^{2} ( x ) - 5 s i n ^{2} ( x ) c o s ( x )}{c o s ( x ) s i n ( x )} x = 0 + + 0 Sin definir 0 < x < \frac{π}{2} + + + + x = \frac{π}{2} + 0 + Sin definir \frac{π}{2} < x < π + - + - x = π + - 0 Sin definir π < x < \frac{3 π}{2} + - - + x = \frac{3 π}{2} + 0 - Sin definir \frac{3 π}{2} < x < 2 π + + - - x = 2 π + + 0 Sin definir

Identificar los intervalos que cumplen la condición:

< 0

\frac{π}{2} < x < π or \frac{3 π}{2} < x < 2 π

Utilizar la periodicidad de

3 tan (x) + cot (x) - 5 sin (x)

\frac{π}{2} + 2 πn < x < π + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

Ejemplos populares

sin(x)-1/2 sqrt(3)<0

sin (x) - \frac{1}{2} 3 < 0

cot^2(x)>= 1/3

cot^{2} (x) \geq \frac{1}{3}

2cos(x)>cos(2x)

2 cos (x) > cos (2 x)

tan^2(x)<1

tan^{2} (x) < 1

2sin^2(x)+3sin(x)+1>0

2 sin^{2} (x) + 3 sin (x) + 1 > 0