8.398. $\left\{\begin{array}{l}x-y=-\frac{1}{3}, \\ \cos ^{2} \pi x-\sin ^{2} \pi y=\frac{1}{2} .\end{array}\right.$
## Solution.

Let's rewrite the second equation of the system as

$$
\begin{aligned}
& \frac{1}{2}(1+\cos 2 \pi x)-\frac{1}{2}(1-\cos 2 \pi y)=\frac{1}{2}, \cos 2 \pi x+\cos 2 \pi y=1 \Leftrightarrow \\
& \Leftrightarrow 2 \cos \frac{2 \pi x+2 \pi y}{2} \cos \frac{2 \pi x-2 \pi y}{2}=1, 2 \cos \pi(x+y) \cos \pi(x-y)=1 \Leftrightarrow \\
& \Leftrightarrow 2 \cos \pi(x+y) \cos \left(-\frac{\pi}{3}\right)=1, 2 \cos \pi(x+y) \cdot \frac{1}{2}=1, \cos \pi(x+y)=1 \\
& \pi(x+y)=2 \pi k, x+y=2 k, k \in \mathbb{Z}
\end{aligned}
$$

Thus, we obtain

$\left\{\begin{array}{l}x-y=-\frac{1}{3}, \\ x+y=2 k, k \in \mathbb{Z} .\end{array} \Rightarrow x=k-\frac{1}{6}, y=k+\frac{1}{6}\right.$, where $k \in \mathbb{Z}$.

Answer: $x=k-\frac{1}{6}, y=k+\frac{1}{6}$, where $k \in \mathbb{Z}$.