8.397. $\left\{\begin{array}{l}\sin x \cos y=0.25, \\ \sin y \cos x=0.75\end{array}\right.$
## Solution.

Let's add and subtract the first and second equations of the system. We get $\left\{\begin{array}{l}\sin x \cos y+\sin y \cos x=1, \\ \sin x \cos y-\sin y \cos x=-\frac{1}{2},\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin (x+y)=1, \\ \sin (x-y)=-\frac{1}{2},\end{array} \Leftrightarrow\right.\right.$ $\Leftrightarrow\left\{\begin{array}{l}x+y=\frac{\pi}{2}+2 \pi k_{1}, \\ x-y=(-1)^{k_{2}} \frac{\pi}{6}+\pi k_{2},\end{array} k_{1}\right.$ and $k_{2} \in Z \Rightarrow$ $x_{1}=\frac{\pi}{6}+\pi\left(k_{1}-k_{2}\right), \quad y_{1}=\frac{\pi}{3}+\pi\left(k_{1}+k_{2}\right) ; x_{2}=-\frac{\pi}{6}+\pi\left(k_{1}-k_{2}\right)$, $y_{2}=\frac{2}{3} \pi+\pi\left(k_{1}+k_{2}\right) ; k_{1}$ and $k_{2} \in Z$.

Answer: $x_{1}=\frac{\pi}{6}+\pi\left(k_{1}-k_{2}\right), y_{1}=\frac{\pi}{3}+\pi\left(k_{1}+k_{2}\right)$;

$$
x_{2}=-\frac{\pi}{6}+\pi\left(k_{1}-k_{2}\right), y_{2}=\frac{2}{3} \pi+\pi\left(k_{1}+k_{2}\right) ; \text { where } k_{1} \text { and } k_{2} \in Z
$$