8.394. $\left\{\begin{array}{l}\sin x+\cos y=0, \\ \sin ^{2} x+\cos ^{2} y=\frac{1}{2} .\end{array}\right.$
## Solution.

Let's write the given system of equations in the form

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ \sin x + \cos y = 0 , } \\
{ (\sin x + \cos y)^2 - 2 \sin x \cos y = \frac { 1 } { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\sin x + \cos y = 0, \\
\sin x \cos y = -\frac{1}{4}
\end{array} \Rightarrow\right.\right. \\
& \Rightarrow \cos y = -\sin x, \sin x \cdot (-\sin x) = -\frac{1}{4}, \sin^2 x = \frac{1}{4}, \sin x = \pm \frac{1}{2}
\end{aligned}
$$

Thus, the given system is equivalent to the following system of two systems of equations:
a) $\left\{\begin{array}{l}\sin x + \cos y = 0 \\ \sin x = -\frac{1}{2}\end{array}\right.$

b) $\left\{\begin{array}{l}\sin x + \cos y = 0 \\ \sin x = \frac{1}{2}\end{array}\right.$

Solving these systems, we find

$$
\left\{\begin{array} { l } 
{ x _ { 1 } = ( - 1 ) ^ { k + 1 } \frac { \pi } { 6 } + \pi k , } \\
{ y _ { 1 } = \pm \frac { \pi } { 3 } + 2 \pi n , }
\end{array} \left\{\begin{array}{l}
x_{2}=(-1)^{k} \frac{\pi}{6}+\pi k \\
y_{2}= \pm \frac{2}{3} \pi + 2 \pi n,
\end{array} k, n \in \mathbb{Z}\right.\right.
$$

Answer: $x_{1}=(-1)^{k+1} \frac{\pi}{6}+\pi k, y_{1}= \pm \frac{\pi}{3}+2 \pi n$,

$$
x_{2}=(-1)^{k} \frac{\pi}{6}+\pi k, \quad y_{2}= \pm \frac{2}{3} \pi + 2 \pi n, \text { where } k \text { and } n \in \mathbb{Z}
$$