Тип 13 № 11527

$$4\sin^3x + 4\sqrt{3}(1 - \sin^2x) + 3\sin x - 4\sqrt{3} = 0$$ $$4\sin^3x + 4\sqrt{3} - 4\sqrt{3}\sin^2x + 3\sin x - 4\sqrt{3} = 0$$ $$4\sin^3x - 4\sqrt{3}\sin^2x + 3\sin x = 0$$ $$\sin x(4\sin^2x - 4\sqrt{3}\sin x + 3) = 0 \Leftrightarrow \left[ \begin{aligned} &\sin x = 0 \\ &4\sin^2x - 4\sqrt{3}\sin x + 3 = 0 \end{aligned} \right. \Leftrightarrow$$ $$\Leftrightarrow \left[ \begin{aligned} &x = \pi k, k \in Z \\ & (2\sin x - \sqrt3)^2 = 0 \end{aligned} \right. \Leftrightarrow \left[ \begin{aligned} &x = \pi k, k \in Z \\ & \sin x = \frac{\sqrt3}{2} \end{aligned} \right. \Leftrightarrow \left[ \begin{aligned} &x = \pi k\\ &x = \frac{\pi}{3} + 2\pi k \\ & x = \frac{2\pi}{3} + 2\pi k \end{aligned} \right., k \in Z$$

Ответ: $x = \pi k; \ x = \frac{\pi}{3} + 2\pi k; \ x = \frac{2\pi}{3} + 2\pi k, \quad k \in \mathbb{Z}$