8.066. $6 \sin ^{2} x+2 \sin ^{2} 2 x=5$.

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

8.066. $6 \sin ^{2} x+2 \sin ^{2} 2 x=5$.
Solution.

Rewrite this equation as

$$
\begin{aligned}
& 6 \cdot \frac{1}{2}(1-\cos 2 x)+2\left(1-\cos ^{2} 2 x\right)-5=0 \Leftrightarrow 2 \cos ^{2} 2 x+3 \cos 2 x=0 \\
& \cos 2 x(2 \cos 2 x+3)=0
\end{aligned}
$$

Then

$\cos 2 x=0, 2 x=\frac{\pi}{2}+\pi n, x_{1}=\frac{\pi}{4}+\frac{\pi n}{2}=\frac{\pi}{4}(2 n+1), n \in Z$;

$2 \cos 2 x+3 \neq 0$.

Answer: $\quad x=\frac{\pi}{4}(2 n+1), n \in Z$.