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Popular Trigonometria >

2cos(x)-1>= 0

Solução

2 cos (x) - 1 \geq 0

Solução

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

+2

Notação de intervalo

[- \frac{π}{3} + 2 πn, \frac{π}{3} + 2 πn]

Decimal

- 1.04719 \dots + 2 πn \leq x \leq 1.04719 \dots + 2 πn

Passos da solução

2 cos (x) - 1 \geq 0

Mova

1

para o lado direito

2 cos (x) - 1 \geq 0

Adicionar

1

a ambos os lados

2 cos (x) - 1 + 1 \geq 0 + 1

Simplificar

2 cos (x) \geq 1

2 cos (x) \geq 1

Dividir ambos os lados por

2

2 cos (x) \geq 1

Dividir ambos os lados por

2

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

Simplificar

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

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

Para

cos (x) \geq a

, se

- 1 < a < 1

então

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

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

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}

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

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