English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1250>600cos((2pi)/3 (t-1))+1000

Solución

1250 > 600 cos (\frac{2 π}{3} (t - 1)) + 1000

Solución

\frac{3 arccos ( \frac{5}{12} ) + 2 π}{2 π} + 3 n < t < \frac{8 π - 3 arccos ( \frac{5}{12} )}{2 π} + 3 n

Notación de intervalos

(\frac{3 arccos ( \frac{5}{12} ) + 2 π}{2 π} + 3 n, \frac{8 π - 3 arccos ( \frac{5}{12} )}{2 π} + 3 n)

Notación decimal

1.54479 \dots + 3 n < t < 3.45520 \dots + 3 n

Pasos de solución

1250 > 600 cos (\frac{2 π}{3} (t - 1)) + 1000

Intercambiar lados

600 cos (\frac{2 π}{3} (t - 1)) + 1000 < 1250

Mover

1000

al lado derecho

600 cos (\frac{2 π}{3} (t - 1)) + 1000 < 1250

Restar

1000

de ambos lados

600 cos (\frac{2 π}{3} (t - 1)) + 1000 - 1000 < 1250 - 1000

Simplificar

600 cos (\frac{2 π}{3} (t - 1)) < 250

600 cos (\frac{2 π}{3} (t - 1)) < 250

Dividir ambos lados entre

600

600 cos (\frac{2 π}{3} (t - 1)) < 250

Dividir ambos lados entre

600

\frac{600 cos ( \frac{2 π}{3} ( t - 1 ) )}{600} < \frac{250}{600}

Simplificar

cos (\frac{2 π}{3} (t - 1)) < \frac{5}{12}

cos (\frac{2 π}{3} (t - 1)) < \frac{5}{12}

Para

cos (x) < a

, si

- 1 < a \leq 1

entonces

arccos (a) + 2 πn < x < 2 π - arccos (a) + 2 πn

arccos (\frac{5}{12}) + 2 πn < \frac{2 π}{3} (t - 1) < 2 π - arccos (\frac{5}{12}) + 2 πn

a < u < b

entonces

a < u and u < b

arccos (\frac{5}{12}) + 2 πn < \frac{2 π}{3} (t - 1) and \frac{2 π}{3} (t - 1) < 2 π - arccos (\frac{5}{12}) + 2 πn

arccos (\frac{5}{12}) + 2 πn < \frac{2 π}{3} (t - 1) : t > \frac{3 arccos ( \frac{5}{12} ) + 2 π}{2 π} + 3 n

arccos (\frac{5}{12}) + 2 πn < \frac{2 π}{3} (t - 1)

Intercambiar lados

\frac{2 π}{3} (t - 1) > arccos (\frac{5}{12}) + 2 πn

Multiplicar ambos lados por

3

\frac{2 π}{3} (t - 1) > arccos (\frac{5}{12}) + 2 πn

Multiplicar ambos lados por

3

3 \cdot \frac{2 π}{3} (t - 1) > 3 arccos (\frac{5}{12}) + 3 \cdot 2 πn

Simplificar

3 \cdot \frac{2 π}{3} (t - 1) > 3 arccos (\frac{5}{12}) + 3 \cdot 2 πn

Simplificar

3 \cdot \frac{2 π}{3} (t - 1) : 2 π (t - 1)

3 \cdot \frac{2 π}{3} (t - 1)

Multiplicar fracciones:

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

= \frac{2 \cdot 3 π}{3} (t - 1)

Eliminar los terminos comunes:

3

= (t - 1) \cdot 2 π

Simplificar

3 arccos (\frac{5}{12}) + 3 \cdot 2 πn : 3 arccos (\frac{5}{12}) + 6 πn

3 arccos (\frac{5}{12}) + 3 \cdot 2 πn

Multiplicar los numeros:

3 \cdot 2 = 6

= 3 arccos (\frac{5}{12}) + 6 πn

2 π (t - 1) > 3 arccos (\frac{5}{12}) + 6 πn

2 π (t - 1) > 3 arccos (\frac{5}{12}) + 6 πn

2 π (t - 1) > 3 arccos (\frac{5}{12}) + 6 πn

Dividir ambos lados entre

2 π

2 π (t - 1) > 3 arccos (\frac{5}{12}) + 6 πn

Dividir ambos lados entre

2 π

\frac{2 π ( t - 1 )}{2 π} > \frac{3 arccos ( \frac{5}{12} )}{2 π} + \frac{6 πn}{2 π}

Simplificar

\frac{2 π ( t - 1 )}{2 π} > \frac{3 arccos ( \frac{5}{12} )}{2 π} + \frac{6 πn}{2 π}

Simplificar

\frac{2 π ( t - 1 )}{2 π} : t - 1

\frac{2 π ( t - 1 )}{2 π}

Dividir:

\frac{2}{2} = 1

= \frac{π ( t - 1 )}{π}

Eliminar los terminos comunes:

π

= t - 1

Simplificar

\frac{3 arccos ( \frac{5}{12} )}{2 π} + \frac{6 πn}{2 π} : \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

\frac{3 arccos ( \frac{5}{12} )}{2 π} + \frac{6 πn}{2 π}

Cancelar

\frac{6 πn}{2 π} : 3 n

\frac{6 πn}{2 π}

Cancelar

\frac{6 πn}{2 π} : 3 n

\frac{6 πn}{2 π}

Dividir:

\frac{6}{2} = 3

= \frac{3 πn}{π}

Eliminar los terminos comunes:

π

= 3 n

= 3 n

= \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

t - 1 > \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

t - 1 > \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

t - 1 > \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

Mover

1

al lado derecho

t - 1 > \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

Sumar

1

a ambos lados

t - 1 + 1 > \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n + 1

Simplificar

t > \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n + 1

t > \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n + 1

Simplificar

\frac{3 arccos ( \frac{5}{12} )}{2 π} + 1 : \frac{3 arccos ( \frac{5}{12} ) + 2 π}{2 π}

\frac{3 arccos ( \frac{5}{12} )}{2 π} + 1

Convertir a fracción:

1 = \frac{1 \cdot 2 π}{2 π}

= \frac{3 arccos ( \frac{5}{12} )}{2 π} + \frac{1 \cdot 2 π}{2 π}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{3 arccos ( \frac{5}{12} ) + 1 \cdot 2 π}{2 π}

Multiplicar los numeros:

1 \cdot 2 = 2

= \frac{3 arccos ( \frac{5}{12} ) + 2 π}{2 π}

t > \frac{3 arccos ( \frac{5}{12} ) + 2 π}{2 π} + 3 n

\frac{2 π}{3} (t - 1) < 2 π - arccos (\frac{5}{12}) + 2 πn : t < \frac{8 π - 3 arccos ( \frac{5}{12} )}{2 π} + 3 n

\frac{2 π}{3} (t - 1) < 2 π - arccos (\frac{5}{12}) + 2 πn

Multiplicar ambos lados por

3

\frac{2 π}{3} (t - 1) < 2 π - arccos (\frac{5}{12}) + 2 πn

Multiplicar ambos lados por

3

3 \cdot \frac{2 π}{3} (t - 1) < 3 \cdot 2 π - 3 arccos (\frac{5}{12}) + 3 \cdot 2 πn

Simplificar

3 \cdot \frac{2 π}{3} (t - 1) < 3 \cdot 2 π - 3 arccos (\frac{5}{12}) + 3 \cdot 2 πn

Simplificar

3 \cdot \frac{2 π}{3} (t - 1) : 2 π (t - 1)

3 \cdot \frac{2 π}{3} (t - 1)

Multiplicar fracciones:

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

= \frac{2 \cdot 3 π}{3} (t - 1)

Eliminar los terminos comunes:

3

= (t - 1) \cdot 2 π

Simplificar

3 \cdot 2 π - 3 arccos (\frac{5}{12}) + 3 \cdot 2 πn : 6 π - 3 arccos (\frac{5}{12}) + 6 πn

3 \cdot 2 π - 3 arccos (\frac{5}{12}) + 3 \cdot 2 πn

Multiplicar los numeros:

3 \cdot 2 = 6

= 6 π - 3 arccos (\frac{5}{12}) + 6 πn

2 π (t - 1) < 6 π - 3 arccos (\frac{5}{12}) + 6 πn

2 π (t - 1) < 6 π - 3 arccos (\frac{5}{12}) + 6 πn

2 π (t - 1) < 6 π - 3 arccos (\frac{5}{12}) + 6 πn

Dividir ambos lados entre

2 π

2 π (t - 1) < 6 π - 3 arccos (\frac{5}{12}) + 6 πn

Dividir ambos lados entre

2 π

\frac{2 π ( t - 1 )}{2 π} < \frac{6 π}{2 π} - \frac{3 arccos ( \frac{5}{12} )}{2 π} + \frac{6 πn}{2 π}

Simplificar

\frac{2 π ( t - 1 )}{2 π} < \frac{6 π}{2 π} - \frac{3 arccos ( \frac{5}{12} )}{2 π} + \frac{6 πn}{2 π}

Simplificar

\frac{2 π ( t - 1 )}{2 π} : t - 1

\frac{2 π ( t - 1 )}{2 π}

Dividir:

\frac{2}{2} = 1

= \frac{π ( t - 1 )}{π}

Eliminar los terminos comunes:

π

= t - 1

Simplificar

\frac{6 π}{2 π} - \frac{3 arccos ( \frac{5}{12} )}{2 π} + \frac{6 πn}{2 π} : 3 - \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

\frac{6 π}{2 π} - \frac{3 arccos ( \frac{5}{12} )}{2 π} + \frac{6 πn}{2 π}

Cancelar

\frac{6 π}{2 π} : 3

\frac{6 π}{2 π}

Cancelar

\frac{6 π}{2 π} : 3

\frac{6 π}{2 π}

Dividir:

\frac{6}{2} = 3

= \frac{3 π}{π}

Eliminar los terminos comunes:

π

= 3

= 3

= 3 - \frac{3 arccos ( \frac{5}{12} )}{2 π} + \frac{6 πn}{2 π}

Cancelar

\frac{6 πn}{2 π} : 3 n

\frac{6 πn}{2 π}

Cancelar

\frac{6 πn}{2 π} : 3 n

\frac{6 πn}{2 π}

Dividir:

\frac{6}{2} = 3

= \frac{3 πn}{π}

Eliminar los terminos comunes:

π

= 3 n

= 3 n

= 3 - \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

t - 1 < 3 - \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

t - 1 < 3 - \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

t - 1 < 3 - \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

Mover

1

al lado derecho

t - 1 < 3 - \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n

Sumar

1

a ambos lados

t - 1 + 1 < 3 - \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n + 1

Simplificar

t - 1 + 1 < 3 - \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n + 1

Simplificar

t - 1 + 1 : t

t - 1 + 1

Sumar elementos similares:

- 1 + 1 < 0

= t

Simplificar

3 - \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n + 1 : 3 n + 4 - \frac{3 arccos ( \frac{5}{12} )}{2 π}

3 - \frac{3 arccos ( \frac{5}{12} )}{2 π} + 3 n + 1

Sumar:

3 + 1 = 4

= 3 n + 4 - \frac{3 arccos ( \frac{5}{12} )}{2 π}

t < 3 n + 4 - \frac{3 arccos ( \frac{5}{12} )}{2 π}

t < 3 n + 4 - \frac{3 arccos ( \frac{5}{12} )}{2 π}

t < 3 n + 4 - \frac{3 arccos ( \frac{5}{12} )}{2 π}

Simplificar

4 - \frac{3 arccos ( \frac{5}{12} )}{2 π} : \frac{8 π - 3 arccos ( \frac{5}{12} )}{2 π}

4 - \frac{3 arccos ( \frac{5}{12} )}{2 π}

Convertir a fracción:

4 = \frac{4 \cdot 2 π}{2 π}

= \frac{4 \cdot 2 π}{2 π} - \frac{3 arccos ( \frac{5}{12} )}{2 π}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{4 \cdot 2 π - 3 arccos ( \frac{5}{12} )}{2 π}

Multiplicar los numeros:

4 \cdot 2 = 8

= \frac{8 π - 3 arccos ( \frac{5}{12} )}{2 π}

t < \frac{8 π - 3 arccos ( \frac{5}{12} )}{2 π} + 3 n

Combinar los rangos

t > \frac{3 arccos ( \frac{5}{12} ) + 2 π}{2 π} + 3 n and t < \frac{8 π - 3 arccos ( \frac{5}{12} )}{2 π} + 3 n

Mezclar intervalos sobrepuestos

\frac{3 arccos ( \frac{5}{12} ) + 2 π}{2 π} + 3 n < t < \frac{8 π - 3 arccos ( \frac{5}{12} )}{2 π} + 3 n

Ejemplos populares

sec(x)<= 2/(sqrt(3))

sec (x) \leq \frac{2}{3}

tan(a)>1

tan (a) > 1

tan(x)>= sin(2x)

tan (x) \geq sin (2 x)

sqrt(3)tan(x)>1

3 tan (x) > 1

cos((pix)/2)> 1/2

cos (\frac{π x}{2}) > \frac{1}{2}