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Popular Trigonometria >

-1<2cos(x)<1

Solução

- 1 < 2 cos (x) < 1

Solução

\frac{π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn or \frac{4 π}{3} + 2 πn < x < \frac{5 π}{3} + 2 πn

Notação de intervalo

(\frac{π}{3} + 2 πn, \frac{2 π}{3} + 2 πn) \cup (\frac{4 π}{3} + 2 πn, \frac{5 π}{3} + 2 πn)

Decimal

1.04719 \dots + 2 πn < x < 2.09439 \dots + 2 πn or 4.18879 \dots + 2 πn < x < 5.23598 \dots + 2 πn

Passos da solução

- 1 < 2 cos (x) < 1

a < u < b

então

a < u and u < b

- 1 < 2 cos (x) and 2 cos (x) < 1

- 1 < 2 cos (x) : - \frac{2 π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn

- 1 < 2 cos (x)

Trocar lados

2 cos (x) > - 1

Dividir ambos os lados por

2

2 cos (x) > - 1

Dividir ambos os lados por

2

\frac{2 cos ( x )}{2} > \frac{- 1}{2}

Simplificar

cos (x) > - \frac{1}{2}

cos (x) > - \frac{1}{2}

Para

cos (x) > a

, se

- 1 \leq a < 1

então

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

- arccos (- \frac{1}{2}) + 2 πn < x < arccos (- \frac{1}{2}) + 2 πn

Simplificar

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

- arccos (- \frac{1}{2})

Utilizar a seguinte identidade trivial:

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

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{2 π}{3}

Simplificar

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

arccos (- \frac{1}{2})

Utilizar a seguinte identidade trivial:

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

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{2 π}{3}

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

2 cos (x) < 1 : \frac{π}{3} + 2 πn < x < \frac{5 π}{3} + 2 πn

2 cos (x) < 1

Dividir ambos os lados por

2

2 cos (x) < 1

Dividir ambos os lados por

2

\frac{2 cos ( x )}{2} < \frac{1}{2}

Simplificar

cos (x) < \frac{1}{2}

cos (x) < \frac{1}{2}

Para

cos (x) < a

, se

- 1 < a \leq 1

então

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

arccos (\frac{1}{2}) + 2 πn < x < 2 π - arccos (\frac{1}{2}) + 2 πn

Simplificar

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

Utilizar a seguinte identidade trivial:

arccos (\frac{1}{2}) = \frac{π}{3}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{3}

Simplificar

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

2 π - arccos (\frac{1}{2})

Utilizar a seguinte identidade trivial:

arccos (\frac{1}{2}) = \frac{π}{3}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{3}

Simplificar

2 π - \frac{π}{3}

Converter para fração:

2 π = \frac{2 π 3}{3}

= \frac{2 π 3}{3} - \frac{π}{3}

Já que os denominadores são iguais, combinar as frações:

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

= \frac{2 π 3 - π}{3}

2 π 3 - π = 5 π

2 π 3 - π

Multiplicar os números:

2 \cdot 3 = 6

= 6 π - π

Somar elementos similares:

6 π - π = 5 π

= 5 π

= \frac{5 π}{3}

= \frac{5 π}{3}

\frac{π}{3} + 2 πn < x < \frac{5 π}{3} + 2 πn

Combinar os intervalos

- \frac{2 π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn and \frac{π}{3} + 2 πn < x < \frac{5 π}{3} + 2 πn

Junte intervalos que se sobrepoem

\frac{π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn or \frac{4 π}{3} + 2 πn < x < \frac{5 π}{3} + 2 πn

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