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Popular Trigonometria >

13<2.5cos(((2pi)/(365))(x+9))+12

Solução

13 < 2.5 cos ((\frac{2 π}{365}) (x + 9)) + 12

Solução

\frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n < x < \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

Notação de intervalo

(\frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n, \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n)

Decimal

- 76.34434 \dots + 365 n < x < 58.34434 \dots + 365 n

Passos da solução

13 < 2.5 cos (\frac{2 π}{365} (x + 9)) + 12

Trocar lados

2.5 cos (\frac{2 π}{365} (x + 9)) + 12 > 13

Multiplicar ambos os lados por

10

2.5 cos (\frac{2 π}{365} (x + 9)) + 12 > 13

To eliminate decimal points, multiply by 10 for every digit after the decimal pointThere is one digit to the right of the decimal point, therefore multiply by

10

2.5 cos (\frac{2 π}{365} (x + 9)) \cdot 10 + 12 \cdot 10 > 13 \cdot 10

Simplificar

25 cos (\frac{2 π}{365} (x + 9)) + 120 > 130

25 cos (\frac{2 π}{365} (x + 9)) + 120 > 130

Mova

120

para o lado direito

25 cos (\frac{2 π}{365} (x + 9)) + 120 > 130

Subtrair

120

de ambos os lados

25 cos (\frac{2 π}{365} (x + 9)) + 120 - 120 > 130 - 120

Simplificar

25 cos (\frac{2 π}{365} (x + 9)) > 10

25 cos (\frac{2 π}{365} (x + 9)) > 10

Dividir ambos os lados por

25

25 cos (\frac{2 π}{365} (x + 9)) > 10

Dividir ambos os lados por

25

\frac{25 cos ( \frac{2 π}{365} ( x + 9 ) )}{25} > \frac{10}{25}

Simplificar

cos (\frac{2 π}{365} (x + 9)) > \frac{2}{5}

cos (\frac{2 π}{365} (x + 9)) > \frac{2}{5}

Para

cos (x) > a

, se

- 1 \leq a < 1

então

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

- arccos (\frac{2}{5}) + 2 πn < \frac{2 π}{365} (x + 9) < arccos (\frac{2}{5}) + 2 πn

a < u < b

então

a < u and u < b

- arccos (\frac{2}{5}) + 2 πn < \frac{2 π}{365} (x + 9) and \frac{2 π}{365} (x + 9) < arccos (\frac{2}{5}) + 2 πn

- arccos (\frac{2}{5}) + 2 πn < \frac{2 π}{365} (x + 9) : x > \frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

- arccos (\frac{2}{5}) + 2 πn < \frac{2 π}{365} (x + 9)

Trocar lados

\frac{2 π}{365} (x + 9) > - arccos (\frac{2}{5}) + 2 πn

Multiplicar ambos os lados por

365

\frac{2 π}{365} (x + 9) > - arccos (\frac{2}{5}) + 2 πn

Multiplicar ambos os lados por

365

365 \cdot \frac{2 π}{365} (x + 9) > - 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

Simplificar

365 \cdot \frac{2 π}{365} (x + 9) > - 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

Simplificar

365 \cdot \frac{2 π}{365} (x + 9) : 2 π (x + 9)

365 \cdot \frac{2 π}{365} (x + 9)

Multiplicar frações:

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

= \frac{2 \cdot 365 π}{365} (x + 9)

Eliminar o fator comum:

365

= (x + 9) \cdot 2 π

Simplificar

- 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn : - 365 arccos (\frac{2}{5}) + 730 πn

- 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

Multiplicar os números:

365 \cdot 2 = 730

= - 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) > - 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) > - 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) > - 365 arccos (\frac{2}{5}) + 730 πn

Dividir ambos os lados por

2 π

2 π (x + 9) > - 365 arccos (\frac{2}{5}) + 730 πn

Dividir ambos os lados por

2 π

\frac{2 π ( x + 9 )}{2 π} > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

Simplificar

\frac{2 π ( x + 9 )}{2 π} > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

Simplificar

\frac{2 π ( x + 9 )}{2 π} : x + 9

\frac{2 π ( x + 9 )}{2 π}

Dividir:

\frac{2}{2} = 1

= \frac{π ( x + 9 )}{π}

Eliminar o fator comum:

π

= x + 9

Simplificar

- \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π} : - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

- \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

Cancelar

\frac{730 πn}{2 π} : 365 n

\frac{730 πn}{2 π}

Cancelar

\frac{730 πn}{2 π} : 365 n

\frac{730 πn}{2 π}

Dividir:

\frac{730}{2} = 365

= \frac{365 πn}{π}

Eliminar o fator comum:

π

= 365 n

= 365 n

= - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

Mova

9

para o lado direito

x + 9 > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

Subtrair

9

de ambos os lados

x + 9 - 9 > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

Simplificar

x > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

x > - \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

Simplificar

- \frac{365 arccos ( \frac{2}{5} )}{2 π} - 9 : \frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π}

- \frac{365 arccos ( \frac{2}{5} )}{2 π} - 9

Converter para fração:

9 = \frac{9 \cdot 2 π}{2 π}

= - \frac{365 arccos ( \frac{2}{5} )}{2 π} - \frac{9 \cdot 2 π}{2 π}

Já que os denominadores são iguais, combinar as frações:

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

= \frac{- 365 arccos ( \frac{2}{5} ) - 9 \cdot 2 π}{2 π}

Multiplicar os números:

9 \cdot 2 = 18

= \frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π}

x > \frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

\frac{2 π}{365} (x + 9) < arccos (\frac{2}{5}) + 2 πn : x < \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

\frac{2 π}{365} (x + 9) < arccos (\frac{2}{5}) + 2 πn

Multiplicar ambos os lados por

365

\frac{2 π}{365} (x + 9) < arccos (\frac{2}{5}) + 2 πn

Multiplicar ambos os lados por

365

365 \cdot \frac{2 π}{365} (x + 9) < 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

Simplificar

365 \cdot \frac{2 π}{365} (x + 9) < 365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

Simplificar

365 \cdot \frac{2 π}{365} (x + 9) : 2 π (x + 9)

365 \cdot \frac{2 π}{365} (x + 9)

Multiplicar frações:

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

= \frac{2 \cdot 365 π}{365} (x + 9)

Eliminar o fator comum:

365

= (x + 9) \cdot 2 π

Simplificar

365 arccos (\frac{2}{5}) + 365 \cdot 2 πn : 365 arccos (\frac{2}{5}) + 730 πn

365 arccos (\frac{2}{5}) + 365 \cdot 2 πn

Multiplicar os números:

365 \cdot 2 = 730

= 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) < 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) < 365 arccos (\frac{2}{5}) + 730 πn

2 π (x + 9) < 365 arccos (\frac{2}{5}) + 730 πn

Dividir ambos os lados por

2 π

2 π (x + 9) < 365 arccos (\frac{2}{5}) + 730 πn

Dividir ambos os lados por

2 π

\frac{2 π ( x + 9 )}{2 π} < \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

Simplificar

\frac{2 π ( x + 9 )}{2 π} < \frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

Simplificar

\frac{2 π ( x + 9 )}{2 π} : x + 9

\frac{2 π ( x + 9 )}{2 π}

Dividir:

\frac{2}{2} = 1

= \frac{π ( x + 9 )}{π}

Eliminar o fator comum:

π

= x + 9

Simplificar

\frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π} : \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

\frac{365 arccos ( \frac{2}{5} )}{2 π} + \frac{730 πn}{2 π}

Cancelar

\frac{730 πn}{2 π} : 365 n

\frac{730 πn}{2 π}

Cancelar

\frac{730 πn}{2 π} : 365 n

\frac{730 πn}{2 π}

Dividir:

\frac{730}{2} = 365

= \frac{365 πn}{π}

Eliminar o fator comum:

π

= 365 n

= 365 n

= \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

x + 9 < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

Mova

9

para o lado direito

x + 9 < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n

Subtrair

9

de ambos os lados

x + 9 - 9 < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

Simplificar

x < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

x < \frac{365 arccos ( \frac{2}{5} )}{2 π} + 365 n - 9

Simplificar

\frac{365 arccos ( \frac{2}{5} )}{2 π} - 9 : \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π}

\frac{365 arccos ( \frac{2}{5} )}{2 π} - 9

Converter para fração:

9 = \frac{9 \cdot 2 π}{2 π}

= \frac{365 arccos ( \frac{2}{5} )}{2 π} - \frac{9 \cdot 2 π}{2 π}

Já que os denominadores são iguais, combinar as frações:

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

= \frac{365 arccos ( \frac{2}{5} ) - 9 \cdot 2 π}{2 π}

Multiplicar os números:

9 \cdot 2 = 18

= \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π}

x < \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

Combinar os intervalos

x > \frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n and x < \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

Junte intervalos que se sobrepoem

\frac{- 365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n < x < \frac{365 arccos ( \frac{2}{5} ) - 18 π}{2 π} + 365 n

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