8.385. $3\left(\log _{2} \sin x\right)^{2}+\log _{2}(1-\cos 2 x)=2$.
## Solution.

Domain of definition: $\sin x>0$.

From the condition we have

$$
\begin{aligned}
& 3\left(\log _{2} \sin x\right)^{2}+\log _{2}\left(1-1+2 \sin ^{2} x\right)-2=0 \Leftrightarrow \\
& \Leftrightarrow 3\left(\log _{2} \sin x\right)^{2}+\log _{2}\left(2 \sin ^{2} x\right)-2=0 \Leftrightarrow \\
& \Leftrightarrow 3\left(\log _{2} \sin x\right)^{2}+2\left(\log _{2} \sin x\right)-1=0
\end{aligned}
$$

Solving this equation as a quadratic in $\log _{2} \sin x$, we get $\log _{2} \sin x=3, \sin x=8, \varnothing$

or

$\log _{2} \sin x=-1, \sin x=\frac{1}{2}, x_{2}=(-1)^{k} \frac{\pi}{6}+\pi k, k \in Z$.

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