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Popular Trigonometria >

cos(x)+1/2 >0,0pi<= x<= 2pi

Solução

cos (x) + \frac{1}{2} > 0, 0 π \leq x \leq 2 π

Solução

0 π \leq x < \frac{2 π}{3} or 2 π - \frac{2 π}{3} < x \leq 2 π

+2

Notação de intervalo

[0 π, \frac{2 π}{3}) \cup (2 π - \frac{2 π}{3}, 2 π]

Decimal

0 \leq x < 2.09439 \dots or 4.18879 \dots < x \leq 6.28318 \dots

Passos da solução

cos (x) + \frac{1}{2} > 0, 0 π \leq x \leq 2 π

Mova

\frac{1}{2}

para o lado direito

cos (x) + \frac{1}{2} > 0

Subtrair

\frac{1}{2}

de ambos os lados

cos (x) + \frac{1}{2} - \frac{1}{2} > 0 - \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

Combinar os intervalos

- \frac{2 π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn and 0 π \leq x \leq 2 π

0 π \leq x < \frac{2 π}{3} or 2 π - \frac{2 π}{3} < x \leq 2 π

Gráfico

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sin (x) < \frac{1}{3}

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