8.404. $\left\{\begin{array}{l}\cos ^{2} x+\cos ^{2} y=0.25, \\ x+y=\frac{5 \pi}{6} .\end{array}\right.$
## Solution.

Let's write the first equation as

$$
\frac{1}{2}(1+\cos 2 x)+\frac{1}{2}(1+\cos 2 y)=0.25, \cos 2 x+\cos 2 y=-\frac{3}{2} \Leftrightarrow
$$

$\Leftrightarrow 2 \cos (x+y) \cos (x-y)=-\frac{3}{2} \Leftrightarrow 2 \cos \frac{5 \pi}{6} \cos (x-y)=-\frac{3}{2}$, $\cos (x-y)=\frac{\sqrt{3}}{2}, x-y= \pm \frac{\pi}{6}+2 \pi k, k \in Z$.

From this, the original system becomes

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ x - y = \pm \frac { \pi } { 6 } + 2 \pi k , } \\
{ x + y = \frac { 5 \pi } { 6 } , }
\end{array} k \in Z \Leftrightarrow \left[\begin{array}{l}
\left\{\begin{array}{l}
x-y=\frac{\pi}{6}+2 \pi k, \\
x+y=\frac{5 \pi}{6} ; \\
\left\{\begin{array}{l}
x-y=-\frac{\pi}{6}+2 \pi k, \\
x+y=\frac{5 \pi}{6}
\end{array}\right.
\end{array} k \in Z \Leftrightarrow\right. \\
\Leftrightarrow\left\{\begin{array} { l } 
{ x _ { 1 } = \frac { \pi } { 2 } + \pi k = \frac { \pi } { 2 } ( 2 k + 1 ) , } \\
{ y _ { 1 } = \frac { \pi } { 3 } - \pi k = \frac { \pi } { 3 } ( 1 - 3 k ) ; }
\end{array} \left\{\begin{array}{l}
x_{2}=\frac{\pi}{3}+\pi k, \\
y_{2}=\frac{\pi}{2}-\pi k,
\end{array}, k \in Z .\right.\right.
\end{array}\right.\right.
\end{aligned}
$$

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

$$
x_{2}=\frac{\pi}{3}(3 k+1), \quad y_{2}=\frac{\pi}{2}(1-2 k), k \in Z
$$