4. Solve the equation $\frac{\cos x}{\sqrt{3}}-\sqrt{\frac{1-\cos 2 x-2 \sin ^{3} x}{6 \sin x-2}}=0$.

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# Solution:

$$
\frac{\cos x}{\sqrt{3}}=\sqrt{\frac{1-\cos 2 x-2 \sin ^{3} x}{6 \sin x-2}} . \text { Given } \cos x \geq 0 \text {, we square both sides of the equation }
$$

$\frac{\cos ^{2} x}{3}=\frac{1-\cos 2 x-2 \sin ^{3} x}{6 \sin x-2} \Leftrightarrow \frac{\cos ^{2} x}{3}=\frac{2 \sin ^{2} x-2 \sin ^{3} x}{6 \sin x-2} \Leftrightarrow \frac{1-\sin ^{2} x}{3}=\frac{\sin ^{2} x(1-\sin x)}{3 \sin x-1}$ $\Leftrightarrow \frac{(1-\sin x)(1+\sin x)}{3}=\frac{\sin ^{2} x(1-\sin x)}{3 \sin x-1} \Leftrightarrow$

1) $\sin x=1 \quad x=\frac{\pi}{2}+2 \pi n, n \in Z$ or 2) $\frac{1+\sin x}{3}=\frac{\sin ^{2} x}{3 \sin x-1} \Leftrightarrow$ $\sin x \neq \frac{1}{3}, \quad(1+\sin x)(3 \sin x-1)=3 \sin ^{2} x \Leftrightarrow \sin x=\frac{1}{2} \Leftrightarrow x=(-1)^{n} \frac{\pi}{6}+\pi n, n \in Z$. Considering the condition $\cos x \geq 0$, we finally have $x=\frac{\pi}{2}+2 \pi n, x=\frac{\pi}{6}+2 \pi n, n \in Z$.

Answer: $x=\frac{\pi}{2}+2 \pi n, x=\frac{\pi}{6}+2 \pi n, n \in Z$.