8.162. $2(1-\cos 2 x)=\sqrt{3} \tan x$.
Solution.

Domain of definition: $\cos x \neq 0$.

From the condition

$$
2(1-\cos 2 x)-\frac{\sqrt{3}(1-\cos 2 x)}{\sin 2 x}=0,(1-\cos 2 x) \cdot\left(2-\frac{\sqrt{3}}{\sin 2 x}\right)=0
$$

## Hence

1) $1-\cos 2 x=0, \cos 2 x=1,2 x=2 \pi k, x_{1}=\pi k, k \in Z$
2) $2-\frac{\sqrt{3}}{\sin 2 x}=0, \sin 2 x=\frac{\sqrt{3}}{2}, 2 x=(-1)^{n} \frac{\pi}{3}+\pi n, x_{2}=(-1)^{n} \frac{\pi}{6}+\frac{\pi n}{2}$, $n \in Z$.

Answer: $x_{1}=\pi k, x_{2}=(-1)^{n} \frac{\pi}{6}+\frac{\pi n}{2}, k, n \in Z$.