8.396. $\left\{\begin{array}{l}x-y=\frac{5 \pi}{3} \\ \sin x=2 \sin y .\end{array}\right.$
## Solution.

From the condition we have

$$
\begin{aligned}
& x=\frac{5 \pi}{3}+y, \sin \left(\frac{5 \pi}{3}+y\right)-2 \sin y=0,-\sin \left(\frac{\pi}{3}-y\right)-2 \sin y=0 \Leftrightarrow \\
& \Leftrightarrow-\sin \frac{\pi}{3} \cos y+\cos \frac{\pi}{3} \sin y-2 \sin y=0 \Leftrightarrow \sqrt{3} \cos y+3 \sin y=0 \Leftrightarrow \\
& \Leftrightarrow \operatorname{tg} y=-\frac{1}{\sqrt{3}}, y=-\frac{\pi}{6}+\pi k=\frac{\pi}{6}(6 k-1), k \in Z . \\
& \text { Then } x=\frac{5 \pi}{3}-\frac{\pi}{6}+\pi k=\frac{3 \pi}{2}+\pi k=\frac{\pi}{2}(2 k+3), k \in Z
\end{aligned}
$$

Answer: $x=\frac{\pi}{2}(2 k+3), y=\frac{\pi}{6}(6 k-1)$, where $k \in Z$.