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Popular Trigonometria >

cos(x/7)<= 1/2

Solução

cos (\frac{x}{7}) \leq \frac{1}{2}

Solução

\frac{7 π}{3} + 14 πn \leq x \leq \frac{35 π}{3} + 14 πn

Notação de intervalo

[\frac{7 π}{3} + 14 πn, \frac{35 π}{3} + 14 πn]

Decimal

7.33038 \dots + 14 πn \leq x \leq 36.65191 \dots + 14 πn

Passos da solução

cos (\frac{x}{7}) \leq \frac{1}{2}

Para

cos (x) \leq a

, se

- 1 < a < 1

então

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

arccos (\frac{1}{2}) + 2 πn \leq \frac{x}{7} \leq 2 π - arccos (\frac{1}{2}) + 2 πn

a \leq u \leq b

então

a \leq u and u \leq b

arccos (\frac{1}{2}) + 2 πn \leq \frac{x}{7} and \frac{x}{7} \leq 2 π - arccos (\frac{1}{2}) + 2 πn

arccos (\frac{1}{2}) + 2 πn \leq \frac{x}{7} : x \geq \frac{7 π}{3} + 14 πn

arccos (\frac{1}{2}) + 2 πn \leq \frac{x}{7}

Trocar lados

\frac{x}{7} \geq arccos (\frac{1}{2}) + 2 πn

Simplificar

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

arccos (\frac{1}{2}) + 2 πn

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} + 2 πn

\frac{x}{7} \geq \frac{π}{3} + 2 πn

Multiplicar ambos os lados por

7

\frac{x}{7} \geq \frac{π}{3} + 2 πn

Multiplicar ambos os lados por

7

\frac{7 x}{7} \geq 7 \cdot \frac{π}{3} + 7 \cdot 2 πn

Simplificar

\frac{7 x}{7} \geq 7 \cdot \frac{π}{3} + 7 \cdot 2 πn

Simplificar

\frac{7 x}{7} : x

\frac{7 x}{7}

Dividir:

\frac{7}{7} = 1

= x

Simplificar

7 \cdot \frac{π}{3} + 7 \cdot 2 πn : \frac{7 π}{3} + 14 πn

7 \cdot \frac{π}{3} + 7 \cdot 2 πn

Multiplicar

7 \cdot \frac{π}{3} : \frac{7 π}{3}

7 \cdot \frac{π}{3}

Multiplicar frações:

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

= \frac{π 7}{3}

= \frac{7 π}{3} + 7 \cdot 2 πn

Multiplicar os números:

7 \cdot 2 = 14

= \frac{7 π}{3} + 14 πn

x \geq \frac{7 π}{3} + 14 πn

x \geq \frac{7 π}{3} + 14 πn

x \geq \frac{7 π}{3} + 14 πn

\frac{x}{7} \leq 2 π - arccos (\frac{1}{2}) + 2 πn : x \leq \frac{35 π}{3} + 14 πn

\frac{x}{7} \leq 2 π - arccos (\frac{1}{2}) + 2 πn

Simplificar

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

2 π - arccos (\frac{1}{2}) + 2 πn

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} + 2 πn

\frac{x}{7} \leq 2 π - \frac{π}{3} + 2 πn

Multiplicar ambos os lados por

7

\frac{x}{7} \leq 2 π - \frac{π}{3} + 2 πn

Multiplicar ambos os lados por

7

\frac{7 x}{7} \leq 7 \cdot 2 π - 7 \cdot \frac{π}{3} + 7 \cdot 2 πn

Simplificar

\frac{7 x}{7} \leq 7 \cdot 2 π - 7 \cdot \frac{π}{3} + 7 \cdot 2 πn

Simplificar

\frac{7 x}{7} : x

\frac{7 x}{7}

Dividir:

\frac{7}{7} = 1

= x

Simplificar

7 \cdot 2 π - 7 \cdot \frac{π}{3} + 7 \cdot 2 πn : 14 π - \frac{7 π}{3} + 14 πn

7 \cdot 2 π - 7 \cdot \frac{π}{3} + 7 \cdot 2 πn

Multiplicar os números:

7 \cdot 2 = 14

= 14 π - 7 \cdot \frac{π}{3} + 7 \cdot 2 πn

Multiplicar

7 \cdot \frac{π}{3} : \frac{7 π}{3}

7 \cdot \frac{π}{3}

Multiplicar frações:

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

= \frac{π 7}{3}

= 14 π - \frac{7 π}{3} + 7 \cdot 2 πn

Multiplicar os números:

7 \cdot 2 = 14

= 14 π - \frac{7 π}{3} + 14 πn

x \leq 14 π - \frac{7 π}{3} + 14 πn

x \leq 14 π - \frac{7 π}{3} + 14 πn

Simplificar

14 π - \frac{7 π}{3} : \frac{35 π}{3}

14 π - \frac{7 π}{3}

Converter para fração:

14 π = \frac{14 π 3}{3}

= \frac{14 π 3}{3} - \frac{7 π}{3}

Já que os denominadores são iguais, combinar as frações:

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

= \frac{14 π 3 - 7 π}{3}

14 π 3 - 7 π = 35 π

14 π 3 - 7 π

Multiplicar os números:

14 \cdot 3 = 42

= 42 π - 7 π

Somar elementos similares:

42 π - 7 π = 35 π

= 35 π

= \frac{35 π}{3}

x \leq \frac{35 π}{3} + 14 πn

x \leq \frac{35 π}{3} + 14 πn

Combinar os intervalos

x \geq \frac{7 π}{3} + 14 πn and x \leq \frac{35 π}{3} + 14 πn

Junte intervalos que se sobrepoem

\frac{7 π}{3} + 14 πn \leq x \leq \frac{35 π}{3} + 14 πn

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