8.400. $\left\{\begin{array}{l}\sqrt{2} \sin x=\sin y, \\ \sqrt{2} \cos x=\sqrt{3} \cos y .\end{array}\right.$
## Solution.

By squaring both equations of the system, we get

$$
\begin{aligned}
& \left\{\begin{array}{l}
2 \sin ^{2} x=\sin ^{2} y, \\
2 \cos ^{2} x=3 \cos ^{2} y .
\end{array} \Rightarrow 2=\sin ^{2} y+3 \cos ^{2} y, 2 \sin ^{2} y+3\left(1-\sin ^{2} y\right)\right. \\
& \sin ^{2} y=\frac{1}{2}, \sin y= \pm \frac{\sqrt{2}}{2}, y= \pm \frac{\pi}{4}+\pi k_{1}, k_{1} \in Z
\end{aligned}
$$

Then $2 \sin ^{2} x=\sin ^{2} y=\frac{1}{2}, \sin ^{2} x=\frac{1}{4}, \sin x= \pm \frac{1}{2}, x= \pm \frac{\pi}{6}+\pi k_{2}$, $k_{2} \in Z$.

Answer: $x= \pm \frac{\pi}{6}+\pi k_{2}, y= \pm \frac{\pi}{4}+\pi k_{1} ; k_{2}$ and $k_{1} \in Z ; k_{2}$ and $k_{1}$ - numbers of the same parity.