8.370. $\sqrt{\cos ^{2} x+\frac{1}{2}}+\sqrt{\sin ^{2} x+\frac{1}{2}}=2$.
## Solution.

Let

$$
\left\{\begin{array} { l } 
{ \sqrt { \cos ^ { 2 } x + \frac { 1 } { 2 } } = u > 0 , } \\
{ \sqrt { \sin ^ { 2 } x + \frac { 1 } { 2 } } = v > 0 . }
\end{array} \Rightarrow \left\{\begin{array}{l}
\cos ^{2} x+\frac{1}{2}=u^{2}>0, \\
\sin ^{2} x+\frac{1}{2}=v^{2}>0
\end{array} \Rightarrow u^{2}+v^{2}=2\right.\right.
$$

We obtained the following system

$\left\{\begin{array}{l}u+v=2, \\ u^{2}+v^{2}=2,\end{array} \Leftrightarrow\left\{\begin{array}{l}u+v=2, \\ (u+v)^{2}-2 u v=2,\end{array} \Leftrightarrow\left\{\begin{array}{l}u+v=2, \\ u v=1 .\end{array} \Rightarrow u=1, v=1\right.\right.\right.$

Then

$\sqrt{\cos ^{2} x+\frac{1}{2}}=1, \cos ^{2} x+\frac{1}{2}=1, \cos ^{2} x=\frac{1}{2}, \cos x= \pm \frac{\sqrt{2}}{2}$,

$x=\frac{\pi}{4}+\frac{\pi k}{2}=\frac{\pi}{4}(2 k+1), k \in Z$.

Answer: $x=\frac{\pi}{4}(2 k+1), k \in Z$