Решение:
$2 \sin x - 2 \sqrt{3} \sin x - 4 ( 1 - \sin ^{2} x ) - \sqrt{3} + 4 = 0$
$2 \sin x - 2 \sqrt{3} \sin x - 4 + 4 \sin ^{2} x - \sqrt{3} + 4 = 0$
$2 \sin x ( 1 + 2 \sin x ) - \sqrt{3} ( 1 + 2 \sin x ) = 0$
$\left[\begin{matrix} \sin x = - \frac{1}{2} \\ \sin x = \frac{\sqrt{3}}{2} \end{matrix}\right.$
$| x_{1} = - \frac{\pi}{6} + 2 \pi k \\ x_{2} = - \frac{5 \pi}{6} + 2 \pi k \\ x_{3} = \frac{\pi}{3} + 2 \pi k \\ x_{4} = \frac{2 \pi}{3} + 2 \pi k k \in \mathbb{Z}$ 

$[ 2 \pi ; \frac{7 \pi}{2} \left]\right.$

$2 \pi + \frac{\pi}{3} = \frac{7 \pi}{3}$
$3 \pi - \frac{\pi}{3} = \frac{8 \pi}{3}$
$3 \pi + \frac{\pi}{6} = \frac{19 \pi}{6} .$

Ответ: а) $- \frac{\pi}{6} + 2 \pi k , - \frac{5 \pi}{6} + 2 \pi k , \frac{\pi}{3} + 2 \pi k , \frac{2 \pi}{3} + 2 \pi k ,$$k \in \mathbb{Z}$ б) $\frac{7 \pi}{3} , \frac{8 \pi}{3} , \frac{19 \pi}{6}$

Источник: ФИПИ