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受欢迎的三角函数 >

cos(θ)>= (sqrt(3))/2 ,2pi>θ>= 0

解答

cos (θ) \geq \frac{3}{2}, 2 π > θ \geq 0

解答

0 \leq θ \leq \frac{π}{6} or 2 π - \frac{π}{6} \leq θ < 2 π

+2

间隔符号

[0, \frac{π}{6}] \cup [2 π - \frac{π}{6}, 2 π)

十进制

0 \leq θ \leq 0.52359 \dots or 5.75958 \dots \leq θ < 6.28318 \dots

求解步骤

cos (θ) \geq \frac{3}{2}, 0 \leq θ < 2 π

对于

cos (x) \geq a

，若

- 1 < a < 1

，则

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

- arccos (\frac{3}{2}) + 2 πn \leq θ \leq arccos (\frac{3}{2}) + 2 πn

化简

- arccos (\frac{3}{2}) : - \frac{π}{6}

- arccos (\frac{3}{2})

使用以下普通恒等式:

arccos (\frac{3}{2}) = \frac{π}{6}

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}{6}

化简

arccos (\frac{3}{2}) : \frac{π}{6}

arccos (\frac{3}{2})

使用以下普通恒等式:

arccos (\frac{3}{2}) = \frac{π}{6}

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}{6}

- \frac{π}{6} + 2 πn \leq θ \leq \frac{π}{6} + 2 πn

合并区间

- \frac{π}{6} + 2 πn \leq θ \leq \frac{π}{6} + 2 πn and 0 \leq θ < 2 π

0 \leq θ \leq \frac{π}{6} or 2 π - \frac{π}{6} \leq θ < 2 π

作图

流行的例子

sin^2(θ)<= cos^2(θ)

sin^{2} (θ) \leq cos^{2} (θ)

-2+3csc(2y-pi)>= 0

- 2 + 3 csc (2 y - π) \geq 0

sin(x)<sin^2(x)

sin (x) < sin^{2} (x)

tan (x) \leq 0

cos (2 t) > 0