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15cos(pi/(15)x-(2pi)/3)+95<= 105

Solución

15 cos (\frac{π}{15} x - \frac{2 π}{3}) + 95 \leq 105

Solución

\frac{10 π + 15 arccos ( \frac{2}{3} )}{π} + 30 n \leq x \leq \frac{40 π - 15 arccos ( \frac{2}{3} )}{π} + 30 n

Notación de intervalos

[\frac{10 π + 15 arccos ( \frac{2}{3} )}{π} + 30 n, \frac{40 π - 15 arccos ( \frac{2}{3} )}{π} + 30 n]

Notación decimal

14.01580 \dots + 30 n \leq x \leq 35.98419 \dots + 30 n

Pasos de solución

15 cos (\frac{π}{15} x - \frac{2 π}{3}) + 95 \leq 105

Mover

95

al lado derecho

15 cos (\frac{π}{15} x - \frac{2 π}{3}) + 95 \leq 105

Restar

95

de ambos lados

15 cos (\frac{π}{15} x - \frac{2 π}{3}) + 95 - 95 \leq 105 - 95

Simplificar

15 cos (\frac{π}{15} x - \frac{2 π}{3}) \leq 10

15 cos (\frac{π}{15} x - \frac{2 π}{3}) \leq 10

Dividir ambos lados entre

15

15 cos (\frac{π}{15} x - \frac{2 π}{3}) \leq 10

Dividir ambos lados entre

15

\frac{15 cos ( \frac{π}{15} x - \frac{2 π}{3} )}{15} \leq \frac{10}{15}

Simplificar

cos (\frac{π}{15} x - \frac{2 π}{3}) \leq \frac{2}{3}

cos (\frac{π}{15} x - \frac{2 π}{3}) \leq \frac{2}{3}

Para

cos (x) \leq a

, si

- 1 < a < 1

entonces

arccos (a) + 2 πn \leq x \leq 2 π - arccos (a) + 2 πn

arccos (\frac{2}{3}) + 2 πn \leq (\frac{π}{15} x - \frac{2 π}{3}) \leq 2 π - arccos (\frac{2}{3}) + 2 πn

a \leq u \leq b

entonces

a \leq u and u \leq b

arccos (\frac{2}{3}) + 2 πn \leq \frac{π}{15} x - \frac{2 π}{3} and \frac{π}{15} x - \frac{2 π}{3} \leq 2 π - arccos (\frac{2}{3}) + 2 πn

arccos (\frac{2}{3}) + 2 πn \leq \frac{π}{15} x - \frac{2 π}{3} : x \geq \frac{10 π + 15 arccos ( \frac{2}{3} )}{π} + 30 n

arccos (\frac{2}{3}) + 2 πn \leq \frac{π}{15} x - \frac{2 π}{3}

Intercambiar lados

\frac{π}{15} x - \frac{2 π}{3} \geq arccos (\frac{2}{3}) + 2 πn

Mover

\frac{2 π}{3}

al lado derecho

\frac{π}{15} x - \frac{2 π}{3} \geq arccos (\frac{2}{3}) + 2 πn

Sumar

\frac{2 π}{3}

a ambos lados

\frac{π}{15} x - \frac{2 π}{3} + \frac{2 π}{3} \geq arccos (\frac{2}{3}) + 2 πn + \frac{2 π}{3}

Simplificar

\frac{π}{15} x \geq arccos (\frac{2}{3}) + 2 πn + \frac{2 π}{3}

\frac{π}{15} x \geq arccos (\frac{2}{3}) + 2 πn + \frac{2 π}{3}

Multiplicar ambos lados por

15

\frac{π}{15} x \geq arccos (\frac{2}{3}) + 2 πn + \frac{2 π}{3}

Multiplicar ambos lados por

15

15 \cdot \frac{π}{15} x \geq 15 arccos (\frac{2}{3}) + 15 \cdot 2 πn + 15 \cdot \frac{2 π}{3}

Simplificar

15 \cdot \frac{π}{15} x \geq 15 arccos (\frac{2}{3}) + 15 \cdot 2 πn + 15 \cdot \frac{2 π}{3}

Simplificar

15 \cdot \frac{π}{15} x : π x

15 \cdot \frac{π}{15} x

Multiplicar fracciones:

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

= \frac{15 π}{15} x

Eliminar los terminos comunes:

15

= x π

Simplificar

15 arccos (\frac{2}{3}) + 15 \cdot 2 πn + 15 \cdot \frac{2 π}{3} : 15 arccos (\frac{2}{3}) + 30 πn + 10 π

15 arccos (\frac{2}{3}) + 15 \cdot 2 πn + 15 \cdot \frac{2 π}{3}

15 \cdot 2 πn = 30 πn

15 \cdot 2 πn

Multiplicar los numeros:

15 \cdot 2 = 30

= 30 πn

15 \cdot \frac{2 π}{3} = 10 π

15 \cdot \frac{2 π}{3}

Multiplicar fracciones:

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

= \frac{2 π 15}{3}

Multiplicar los numeros:

2 \cdot 15 = 30

= \frac{30 π}{3}

Dividir:

\frac{30}{3} = 10

= 10 π

= 15 arccos (\frac{2}{3}) + 30 πn + 10 π

π x \geq 15 arccos (\frac{2}{3}) + 30 πn + 10 π

π x \geq 15 arccos (\frac{2}{3}) + 30 πn + 10 π

π x \geq 15 arccos (\frac{2}{3}) + 30 πn + 10 π

Dividir ambos lados entre

π

π x \geq 15 arccos (\frac{2}{3}) + 30 πn + 10 π

Dividir ambos lados entre

π

\frac{π x}{π} \geq \frac{15 arccos ( \frac{2}{3} )}{π} + \frac{30 πn}{π} + \frac{10 π}{π}

Simplificar

\frac{π x}{π} \geq \frac{15 arccos ( \frac{2}{3} )}{π} + \frac{30 πn}{π} + \frac{10 π}{π}

Simplificar

\frac{π x}{π} : x

\frac{π x}{π}

Eliminar los terminos comunes:

π

= x

Simplificar

\frac{15 arccos ( \frac{2}{3} )}{π} + \frac{30 πn}{π} + \frac{10 π}{π} : 10 + 30 n + \frac{15 arccos ( \frac{2}{3} )}{π}

\frac{15 arccos ( \frac{2}{3} )}{π} + \frac{30 πn}{π} + \frac{10 π}{π}

Agrupar términos semejantes

= \frac{10 π}{π} + \frac{30 πn}{π} + \frac{15 arccos ( \frac{2}{3} )}{π}

Cancelar

\frac{10 π}{π} : 10

\frac{10 π}{π}

Eliminar los terminos comunes:

π

= 10

= 10 + \frac{30 πn}{π} + \frac{15 arccos ( \frac{2}{3} )}{π}

Cancelar

\frac{30 πn}{π} : 30 n

\frac{30 πn}{π}

Eliminar los terminos comunes:

π

= 30 n

= 10 + 30 n + \frac{15 arccos ( \frac{2}{3} )}{π}

x \geq 10 + 30 n + \frac{15 arccos ( \frac{2}{3} )}{π}

x \geq 10 + 30 n + \frac{15 arccos ( \frac{2}{3} )}{π}

Simplificar

10 + \frac{15 arccos ( \frac{2}{3} )}{π} : \frac{10 π + 15 arccos ( \frac{2}{3} )}{π}

10 + \frac{15 arccos ( \frac{2}{3} )}{π}

Convertir a fracción:

10 = \frac{10 π}{π}

= \frac{10 π}{π} + \frac{15 arccos ( \frac{2}{3} )}{π}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{10 π + 15 arccos ( \frac{2}{3} )}{π}

x \geq \frac{10 π + 15 arccos ( \frac{2}{3} )}{π} + 30 n

x \geq \frac{10 π + 15 arccos ( \frac{2}{3} )}{π} + 30 n

\frac{π}{15} x - \frac{2 π}{3} \leq 2 π - arccos (\frac{2}{3}) + 2 πn : x \leq \frac{40 π - 15 arccos ( \frac{2}{3} )}{π} + 30 n

\frac{π}{15} x - \frac{2 π}{3} \leq 2 π - arccos (\frac{2}{3}) + 2 πn

Mover

\frac{2 π}{3}

al lado derecho

\frac{π}{15} x - \frac{2 π}{3} \leq 2 π - arccos (\frac{2}{3}) + 2 πn

Sumar

\frac{2 π}{3}

a ambos lados

\frac{π}{15} x - \frac{2 π}{3} + \frac{2 π}{3} \leq 2 π - arccos (\frac{2}{3}) + 2 πn + \frac{2 π}{3}

Simplificar

\frac{π}{15} x \leq 2 π - arccos (\frac{2}{3}) + 2 πn + \frac{2 π}{3}

\frac{π}{15} x \leq 2 π - arccos (\frac{2}{3}) + 2 πn + \frac{2 π}{3}

Multiplicar ambos lados por

15

\frac{π}{15} x \leq 2 π - arccos (\frac{2}{3}) + 2 πn + \frac{2 π}{3}

Multiplicar ambos lados por

15

15 \cdot \frac{π}{15} x \leq 15 \cdot 2 π - 15 arccos (\frac{2}{3}) + 15 \cdot 2 πn + 15 \cdot \frac{2 π}{3}

Simplificar

15 \cdot \frac{π}{15} x \leq 15 \cdot 2 π - 15 arccos (\frac{2}{3}) + 15 \cdot 2 πn + 15 \cdot \frac{2 π}{3}

Simplificar

15 \cdot \frac{π}{15} x : π x

15 \cdot \frac{π}{15} x

Multiplicar fracciones:

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

= \frac{15 π}{15} x

Eliminar los terminos comunes:

15

= x π

Simplificar

15 \cdot 2 π - 15 arccos (\frac{2}{3}) + 15 \cdot 2 πn + 15 \cdot \frac{2 π}{3} : 40 π + 30 πn - 15 arccos (\frac{2}{3})

15 \cdot 2 π - 15 arccos (\frac{2}{3}) + 15 \cdot 2 πn + 15 \cdot \frac{2 π}{3}

15 \cdot 2 π = 30 π

15 \cdot 2 π

Multiplicar los numeros:

15 \cdot 2 = 30

= 30 π

15 \cdot 2 πn = 30 πn

15 \cdot 2 πn

Multiplicar los numeros:

15 \cdot 2 = 30

= 30 πn

15 \cdot \frac{2 π}{3} = 10 π

15 \cdot \frac{2 π}{3}

Multiplicar fracciones:

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

= \frac{2 π 15}{3}

Multiplicar los numeros:

2 \cdot 15 = 30

= \frac{30 π}{3}

Dividir:

\frac{30}{3} = 10

= 10 π

= 30 π - 15 arccos (\frac{2}{3}) + 30 πn + 10 π

Agrupar términos semejantes

= 30 π + 10 π + 30 πn - 15 arccos (\frac{2}{3})

Sumar elementos similares:

30 π + 10 π = 40 π

= 40 π + 30 πn - 15 arccos (\frac{2}{3})

π x \leq 40 π + 30 πn - 15 arccos (\frac{2}{3})

π x \leq 40 π + 30 πn - 15 arccos (\frac{2}{3})

π x \leq 40 π + 30 πn - 15 arccos (\frac{2}{3})

Dividir ambos lados entre

π

π x \leq 40 π + 30 πn - 15 arccos (\frac{2}{3})

Dividir ambos lados entre

π

\frac{π x}{π} \leq \frac{40 π}{π} + \frac{30 πn}{π} - \frac{15 arccos ( \frac{2}{3} )}{π}

Simplificar

\frac{π x}{π} \leq \frac{40 π}{π} + \frac{30 πn}{π} - \frac{15 arccos ( \frac{2}{3} )}{π}

Simplificar

\frac{π x}{π} : x

\frac{π x}{π}

Eliminar los terminos comunes:

π

= x

Simplificar

\frac{40 π}{π} + \frac{30 πn}{π} - \frac{15 arccos ( \frac{2}{3} )}{π} : 40 + 30 n - \frac{15 arccos ( \frac{2}{3} )}{π}

\frac{40 π}{π} + \frac{30 πn}{π} - \frac{15 arccos ( \frac{2}{3} )}{π}

Cancelar

\frac{40 π}{π} : 40

\frac{40 π}{π}

Eliminar los terminos comunes:

π

= 40

= 40 + \frac{30 πn}{π} - \frac{15 arccos ( \frac{2}{3} )}{π}

Cancelar

\frac{30 πn}{π} : 30 n

\frac{30 πn}{π}

Eliminar los terminos comunes:

π

= 30 n

= 40 + 30 n - \frac{15 arccos ( \frac{2}{3} )}{π}

x \leq 40 + 30 n - \frac{15 arccos ( \frac{2}{3} )}{π}

x \leq 40 + 30 n - \frac{15 arccos ( \frac{2}{3} )}{π}

Simplificar

40 - \frac{15 arccos ( \frac{2}{3} )}{π} : \frac{40 π - 15 arccos ( \frac{2}{3} )}{π}

40 - \frac{15 arccos ( \frac{2}{3} )}{π}

Convertir a fracción:

40 = \frac{40 π}{π}

= \frac{40 π}{π} - \frac{15 arccos ( \frac{2}{3} )}{π}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{40 π - 15 arccos ( \frac{2}{3} )}{π}

x \leq \frac{40 π - 15 arccos ( \frac{2}{3} )}{π} + 30 n

x \leq \frac{40 π - 15 arccos ( \frac{2}{3} )}{π} + 30 n

Combinar los rangos

x \geq \frac{10 π + 15 arccos ( \frac{2}{3} )}{π} + 30 n and x \leq \frac{40 π - 15 arccos ( \frac{2}{3} )}{π} + 30 n

Mezclar intervalos sobrepuestos

\frac{10 π + 15 arccos ( \frac{2}{3} )}{π} + 30 n \leq x \leq \frac{40 π - 15 arccos ( \frac{2}{3} )}{π} + 30 n

Ejemplos populares

sin(θ)>0,sec(θ)<0

sin (θ) > 0, sec (θ) < 0

tan^2(x)>= 1

tan^{2} (x) \geq 1

arctan(x)>0

arctan (x) > 0

tan(θ)<0

tan (θ) < 0

sin(x)>-(sqrt(3))/2

sin (x) > - \frac{3}{2}