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sin(x)-1/2 sqrt(3)<0,-pi<= x<= pi

Solução

sin (x) - \frac{1}{2} 3 < 0, - π \leq x \leq π

- π \leq x < \frac{π}{3} or 2 π - \frac{4 π}{3} < x \leq π

Notação de intervalo

[- π, \frac{π}{3}) \cup (2 π - \frac{4 π}{3}, π]

Decimal

- 3.14159 \dots \leq x < 1.04719 \dots or 2.09439 \dots < x \leq 3.14159 \dots

Passos da solução

sin (x) - \frac{1}{2} 3 < 0, - π \leq x \leq π

\frac{1}{2} 3 = \frac{3}{2}

\frac{1}{2} 3

Multiplicar frações:

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

= \frac{1 \cdot 3}{2}

Multiplicar:

1 \cdot 3 = 3

= \frac{3}{2}

sin (x) - \frac{3}{2} < 0

Mova

\frac{3}{2}

para o lado direito

sin (x) - \frac{3}{2} < 0

Adicionar

\frac{3}{2}

a ambos os lados

sin (x) - \frac{3}{2} + \frac{3}{2} < 0 + \frac{3}{2}

Simplificar

sin (x) < \frac{3}{2}

sin (x) < \frac{3}{2}

Para

sin (x) < a

, se

- 1 < a \leq 1

então

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

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

Simplificar

- π - arcsin (\frac{3}{2}) : - \frac{4 π}{3}

- π - arcsin (\frac{3}{2})

Utilizar a seguinte identidade trivial:

arcsin (\frac{3}{2}) = \frac{π}{3}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= - π - \frac{π}{3}

Simplificar

- π - \frac{π}{3}

Converter para fração:

π = \frac{π 3}{3}

= - \frac{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{- π 3 - π}{3}

Somar elementos similares:

- 3 π - π = - 4 π

= \frac{- 4 π}{3}

Aplicar as propriedades das frações:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4 π}{3}

= - \frac{4 π}{3}

Simplificar

arcsin (\frac{3}{2}) : \frac{π}{3}

arcsin (\frac{3}{2})

Utilizar a seguinte identidade trivial:

arcsin (\frac{3}{2}) = \frac{π}{3}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{3}

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

Combinar os intervalos

- \frac{4 π}{3} + 2 πn < x < \frac{π}{3} + 2 πn and - π \leq x \leq π

- π \leq x < \frac{π}{3} or 2 π - \frac{4 π}{3} < x \leq π

sin (x) + cos (x) \geq 1

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

sin (θ) < 0, tan (θ) < 0

cos (x) < sin (2 x)

7.5 cos (\frac{π}{6} (x + 3)) + 10.5 > 13.75