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Popular Trigonometría >

(-1/5)*cos(2 pi/5 (x+1))+1>= 16/15

Solución

(- \frac{1}{5}) \cdot cos (2 \frac{π}{5} (x + 1)) + 1 \geq \frac{16}{15}

Solución

\frac{5 arccos ( - \frac{1}{3} ) - 2 π}{2 π} + 5 n \leq x \leq \frac{8 π - 5 arccos ( - \frac{1}{3} )}{2 π} + 5 n

Notación de intervalos

[\frac{5 arccos ( - \frac{1}{3} ) - 2 π}{2 π} + 5 n, \frac{8 π - 5 arccos ( - \frac{1}{3} )}{2 π} + 5 n]

Decimal

0.52043 \dots + 5 n \leq x \leq 2.47956 \dots + 5 n

Pasos de solución

(- \frac{1}{5}) cos (2 \cdot \frac{π}{5} (x + 1)) + 1 \geq \frac{16}{15}

Desplace

1

a la derecha

(- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1)) + 1 \geq \frac{16}{15}

Restar

1

de ambos lados

(- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1)) + 1 - 1 \geq \frac{16}{15} - 1

Simplificar

(- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1)) + 1 - 1 \geq \frac{16}{15} - 1

Simplificar

(- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1)) + 1 - 1 : (- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1))

(- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1)) + 1 - 1

Sumar elementos similares:

1 - 1 \geq 0

= (- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1))

Simplificar

\frac{16}{15} - 1 : \frac{1}{15}

\frac{16}{15} - 1

Convertir a fracción:

1 = \frac{1 \cdot 15}{15}

= - \frac{1 \cdot 15}{15} + \frac{16}{15}

Ya que los denominadores son iguales, combinar las fracciones:

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

= \frac{- 1 \cdot 15 + 16}{15}

- 1 \cdot 15 + 16 = 1

- 1 \cdot 15 + 16

Multiplicar los numeros:

1 \cdot 15 = 15

= - 15 + 16

Sumar/restar lo siguiente:

- 15 + 16 = 1

= 1

= \frac{1}{15}

(- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1)) \geq \frac{1}{15}

(- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1)) \geq \frac{1}{15}

(- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1)) \geq \frac{1}{15}

Multiplicar ambos lados por

- 1

(- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1)) \geq \frac{1}{15}

Multiplicar ambos lados por -1 (invierte la desigualdad)

(- \frac{1}{5}) cos (2 \frac{π}{5} (x + 1)) (- 1) \leq \frac{1 \cdot ( - 1 )}{15}

Simplificar

\frac{1}{5} cos (2 \frac{π}{5} (x + 1)) \leq - \frac{1}{15}

\frac{1}{5} cos (2 \frac{π}{5} (x + 1)) \leq - \frac{1}{15}

Multiplicar ambos lados por

5

\frac{1}{5} cos (2 \frac{π}{5} (x + 1)) \leq - \frac{1}{15}

Multiplicar ambos lados por

5

5 \cdot \frac{1}{5} cos (2 \frac{π}{5} (x + 1)) \leq 5 (- \frac{1}{15})

Simplificar

5 \cdot \frac{1}{5} cos (2 \frac{π}{5} (x + 1)) \leq 5 (- \frac{1}{15})

Simplificar

5 \cdot \frac{1}{5} cos (2 \frac{π}{5} (x + 1)) : cos (2 \frac{π}{5} (x + 1))

5 \cdot \frac{1}{5} cos (2 \frac{π}{5} (x + 1))

Multiplicar fracciones:

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

= \frac{1 \cdot 5}{5} cos (2 \frac{π}{5} (x + 1))

Eliminar los terminos comunes:

5

= cos (2 \frac{π}{5} (x + 1)) \cdot 1

Multiplicar:

cos (2 \frac{π}{5} (x + 1)) \cdot 1 = cos (2 \frac{π}{5} (x + 1))

= cos (2 \frac{π}{5} (x + 1))

Simplificar

5 (- \frac{1}{15}) : - \frac{1}{3}

5 (- \frac{1}{15})

Quitar los parentesis:

(- a) = - a

= - 5 \cdot \frac{1}{15}

Multiplicar fracciones:

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

= - \frac{1 \cdot 5}{15}

Multiplicar los numeros:

1 \cdot 5 = 5

= - \frac{5}{15}

Eliminar los terminos comunes:

5

= - \frac{1}{3}

cos (2 \frac{π}{5} (x + 1)) \leq - \frac{1}{3}

cos (2 \frac{π}{5} (x + 1)) \leq - \frac{1}{3}

cos (2 \frac{π}{5} (x + 1)) \leq - \frac{1}{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{1}{3}) + 2 πn \leq 2 \cdot \frac{π}{5} (x + 1) \leq 2 π - arccos (- \frac{1}{3}) + 2 πn

a \leq u \leq b

entonces

a \leq u and u \leq b

arccos (- \frac{1}{3}) + 2 πn \leq 2 \cdot \frac{π}{5} (x + 1) and 2 \cdot \frac{π}{5} (x + 1) \leq 2 π - arccos (- \frac{1}{3}) + 2 πn

arccos (- \frac{1}{3}) + 2 πn \leq 2 \cdot \frac{π}{5} (x + 1) : x \geq \frac{5 arccos ( - \frac{1}{3} ) - 2 π}{2 π} + 5 n

arccos (- \frac{1}{3}) + 2 πn \leq 2 \cdot \frac{π}{5} (x + 1)

Intercambiar lados

2 \cdot \frac{π}{5} (x + 1) \geq arccos (- \frac{1}{3}) + 2 πn

Simplificar

2 \cdot \frac{π}{5} : \frac{2 π}{5}

2 \cdot \frac{π}{5}

Multiplicar fracciones:

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

= \frac{π 2}{5}

\frac{2 π}{5} (x + 1) \geq arccos (- \frac{1}{3}) + 2 πn

Multiplicar ambos lados por

5

\frac{2 π}{5} (x + 1) \geq arccos (- \frac{1}{3}) + 2 πn

Multiplicar ambos lados por

5

5 \cdot \frac{2 π}{5} (x + 1) \geq 5 arccos (- \frac{1}{3}) + 5 \cdot 2 πn

Simplificar

5 \cdot \frac{2 π}{5} (x + 1) \geq 5 arccos (- \frac{1}{3}) + 5 \cdot 2 πn

Simplificar

5 \cdot \frac{2 π}{5} (x + 1) : 2 π (x + 1)

5 \cdot \frac{2 π}{5} (x + 1)

Multiplicar fracciones:

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

= \frac{2 \cdot 5 π}{5} (x + 1)

Eliminar los terminos comunes:

5

= (x + 1) \cdot 2 π

Simplificar

5 arccos (- \frac{1}{3}) + 5 \cdot 2 πn : 5 arccos (- \frac{1}{3}) + 10 πn

5 arccos (- \frac{1}{3}) + 5 \cdot 2 πn

Multiplicar los numeros:

5 \cdot 2 = 10

= 5 arccos (- \frac{1}{3}) + 10 πn

2 π (x + 1) \geq 5 arccos (- \frac{1}{3}) + 10 πn

2 π (x + 1) \geq 5 arccos (- \frac{1}{3}) + 10 πn

2 π (x + 1) \geq 5 arccos (- \frac{1}{3}) + 10 πn

Dividir ambos lados entre

2 π

2 π (x + 1) \geq 5 arccos (- \frac{1}{3}) + 10 πn

Dividir ambos lados entre

2 π

\frac{2 π ( x + 1 )}{2 π} \geq \frac{5 arccos ( - \frac{1}{3} )}{2 π} + \frac{10 πn}{2 π}

Simplificar

\frac{2 π ( x + 1 )}{2 π} \geq \frac{5 arccos ( - \frac{1}{3} )}{2 π} + \frac{10 πn}{2 π}

Simplificar

\frac{2 π ( x + 1 )}{2 π} : x + 1

\frac{2 π ( x + 1 )}{2 π}

Dividir:

\frac{2}{2} = 1

= \frac{π ( x + 1 )}{π}

Eliminar los terminos comunes:

π

= x + 1

Simplificar

\frac{5 arccos ( - \frac{1}{3} )}{2 π} + \frac{10 πn}{2 π} : \frac{5 arccos ( - \frac{1}{3} )}{2 π} + 5 n

\frac{5 arccos ( - \frac{1}{3} )}{2 π} + \frac{10 πn}{2 π}

Cancelar

\frac{10 πn}{2 π} : 5 n

\frac{10 πn}{2 π}

Cancelar

\frac{10 πn}{2 π} : 5 n

\frac{10 πn}{2 π}

Dividir:

\frac{10}{2} = 5

= \frac{5 πn}{π}

Eliminar los terminos comunes:

π

= 5 n

= 5 n

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

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

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

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

Desplace

1

a la derecha

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

Restar

1

de ambos lados

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

Simplificar

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

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

Simplificar

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

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

Convertir a fracción:

1 = \frac{1 \cdot 2 π}{2 π}

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

Multiplicar los numeros:

1 \cdot 2 = 2

= \frac{5 arccos ( - \frac{1}{3} ) - 2 π}{2 π}

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

2 \cdot \frac{π}{5} (x + 1) \leq 2 π - arccos (- \frac{1}{3}) + 2 πn : x \leq \frac{8 π - 5 arccos ( - \frac{1}{3} )}{2 π} + 5 n

2 \cdot \frac{π}{5} (x + 1) \leq 2 π - arccos (- \frac{1}{3}) + 2 πn

Simplificar

2 \cdot \frac{π}{5} : \frac{2 π}{5}

2 \cdot \frac{π}{5}

Multiplicar fracciones:

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

= \frac{π 2}{5}

\frac{2 π}{5} (x + 1) \leq 2 π - arccos (- \frac{1}{3}) + 2 πn

Multiplicar ambos lados por

5

\frac{2 π}{5} (x + 1) \leq 2 π - arccos (- \frac{1}{3}) + 2 πn

Multiplicar ambos lados por

5

5 \cdot \frac{2 π}{5} (x + 1) \leq 5 \cdot 2 π - 5 arccos (- \frac{1}{3}) + 5 \cdot 2 πn

Simplificar

5 \cdot \frac{2 π}{5} (x + 1) \leq 5 \cdot 2 π - 5 arccos (- \frac{1}{3}) + 5 \cdot 2 πn

Simplificar

5 \cdot \frac{2 π}{5} (x + 1) : 2 π (x + 1)

5 \cdot \frac{2 π}{5} (x + 1)

Multiplicar fracciones:

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

= \frac{2 \cdot 5 π}{5} (x + 1)

Eliminar los terminos comunes:

5

= (x + 1) \cdot 2 π

Simplificar

5 \cdot 2 π - 5 arccos (- \frac{1}{3}) + 5 \cdot 2 πn : 10 π - 5 arccos (- \frac{1}{3}) + 10 πn

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

Multiplicar los numeros:

5 \cdot 2 = 10

= 10 π - 5 arccos (- \frac{1}{3}) + 10 πn

2 π (x + 1) \leq 10 π - 5 arccos (- \frac{1}{3}) + 10 πn

2 π (x + 1) \leq 10 π - 5 arccos (- \frac{1}{3}) + 10 πn

2 π (x + 1) \leq 10 π - 5 arccos (- \frac{1}{3}) + 10 πn

Dividir ambos lados entre

2 π

2 π (x + 1) \leq 10 π - 5 arccos (- \frac{1}{3}) + 10 πn

Dividir ambos lados entre

2 π

\frac{2 π ( x + 1 )}{2 π} \leq \frac{10 π}{2 π} - \frac{5 arccos ( - \frac{1}{3} )}{2 π} + \frac{10 πn}{2 π}

Simplificar

\frac{2 π ( x + 1 )}{2 π} \leq \frac{10 π}{2 π} - \frac{5 arccos ( - \frac{1}{3} )}{2 π} + \frac{10 πn}{2 π}

Simplificar

\frac{2 π ( x + 1 )}{2 π} : x + 1

\frac{2 π ( x + 1 )}{2 π}

Dividir:

\frac{2}{2} = 1

= \frac{π ( x + 1 )}{π}

Eliminar los terminos comunes:

π

= x + 1

Simplificar

\frac{10 π}{2 π} - \frac{5 arccos ( - \frac{1}{3} )}{2 π} + \frac{10 πn}{2 π} : 5 - \frac{5 arccos ( - \frac{1}{3} )}{2 π} + 5 n

\frac{10 π}{2 π} - \frac{5 arccos ( - \frac{1}{3} )}{2 π} + \frac{10 πn}{2 π}

Cancelar

\frac{10 π}{2 π} : 5

\frac{10 π}{2 π}

Cancelar

\frac{10 π}{2 π} : 5

\frac{10 π}{2 π}

Dividir:

\frac{10}{2} = 5

= \frac{5 π}{π}

Eliminar los terminos comunes:

π

= 5

= 5

= 5 - \frac{5 arccos ( - \frac{1}{3} )}{2 π} + \frac{10 πn}{2 π}

Cancelar

\frac{10 πn}{2 π} : 5 n

\frac{10 πn}{2 π}

Cancelar

\frac{10 πn}{2 π} : 5 n

\frac{10 πn}{2 π}

Dividir:

\frac{10}{2} = 5

= \frac{5 πn}{π}

Eliminar los terminos comunes:

π

= 5 n

= 5 n

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

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

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

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

Desplace

1

a la derecha

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

Restar

1

de ambos lados

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

Simplificar

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

Simplificar

x + 1 - 1 : x

x + 1 - 1

Sumar elementos similares:

1 - 1 \leq 0

= x

Simplificar

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

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

Restar:

5 - 1 = 4

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

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

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

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

Simplificar

4 - \frac{5 arccos ( - \frac{1}{3} )}{2 π} : \frac{8 π - 5 arccos ( - \frac{1}{3} )}{2 π}

4 - \frac{5 arccos ( - \frac{1}{3} )}{2 π}

Convertir a fracción:

4 = \frac{4 \cdot 2 π}{2 π}

= \frac{4 \cdot 2 π}{2 π} - \frac{5 arccos ( - \frac{1}{3} )}{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 - 5 arccos ( - \frac{1}{3} )}{2 π}

Multiplicar los numeros:

4 \cdot 2 = 8

= \frac{8 π - 5 arccos ( - \frac{1}{3} )}{2 π}

x \leq \frac{8 π - 5 arccos ( - \frac{1}{3} )}{2 π} + 5 n

Combinar los rangos

x \geq \frac{5 arccos ( - \frac{1}{3} ) - 2 π}{2 π} + 5 n and x \leq \frac{8 π - 5 arccos ( - \frac{1}{3} )}{2 π} + 5 n

Mezclar intervalos sobrepuestos

\frac{5 arccos ( - \frac{1}{3} ) - 2 π}{2 π} + 5 n \leq x \leq \frac{8 π - 5 arccos ( - \frac{1}{3} )}{2 π} + 5 n

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