8.301. $\cos z + \sin z = \sqrt{1 - 2 \cos^2 z}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}1-2 \cos ^{2} z \geq 0 \\ \cos z+\sin z \geq 0\end{array}\right.$

Square both sides of the equation. We have

$$
\left\{\begin{array}{l}
\cos ^{2} z+2 \cos z \sin z+\sin ^{2} z=1-2 \cos ^{2} z \\
\cos z+\sin z \geq 0
\end{array}\right.
$$

$\Rightarrow 2 \cos z \sin z+2 \cos ^{2} z=0 \Leftrightarrow \cos z(\sin z+\cos z)=0$, from which

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0565.jpg?height=83&width=1163&top_left_y=1081&top_left_x=46)
$=\frac{\pi}{2}(4 k+1), k \in Z$.

2) $\sin z+\cos z=0, \operatorname{tg} z=-1, z_{2}=-\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n-1), n \in Z$.

Answer: $z_{1}=\frac{\pi}{2}(4 k+1) ; z_{2}=\frac{\pi}{4}(4 n-1)$, where $k$ and $n \in Z$.