8.077. $2 \sin z - \cos z = \frac{2}{5}$.
Solution.

By transitioning to the half-argument, we find

$$
\begin{aligned}
& 20 \sin \frac{z}{2} \cos \frac{z}{2} - 5 \left( \cos^2 \frac{z}{2} - \sin^2 \frac{z}{2} \right) - 2 \left( \cos^2 \frac{z}{2} + \sin^2 \frac{z}{2} \right) = 0 \\
& 3 \sin^2 \frac{z}{2} + 20 \sin \frac{z}{2} \cos \frac{z}{2} - 7 \cos^2 \frac{z}{2} = 0 \Leftrightarrow 3 \operatorname{tg}^2 \frac{z}{2} + 20 \operatorname{tg} \frac{z}{2} - 7 = 0
\end{aligned}
$$

Solving this equation as a quadratic in $\operatorname{tg} \frac{z}{2}$, we find $\operatorname{tg} \frac{z}{2} = -7$,
$\frac{z}{2} = -\operatorname{arctg} 7 + \pi k, z_1 = -2 \operatorname{arctg} 7 + 2 \pi k, k \in \mathbb{Z}; \operatorname{tg} \frac{z}{2} = \frac{1}{3}, \frac{\pi}{2} = \operatorname{arctg} \frac{1}{3} + \pi l, z_2 = 2 \operatorname{arctg} \frac{1}{3} + 2 \pi l = 2 \operatorname{arcctg} 3 + 2 \pi l, l \in \mathbb{Z}$.

Answer: $z_1 = -2 \operatorname{arctg} 7 + 2 \pi k; z_2 = 2 \operatorname{arcctg} 3 + 2 \pi l, k, l \in \mathbb{Z}$. 8.078. $\cos \left(\frac{\pi}{2} + 5x\right) + \sin x = 2 \cos 3x$.

## Solution.

From the condition

$(\sin 5x - \sin x) + 2 \cos 3x = 0 \Leftrightarrow 2 \cos \frac{5x + x}{2} \sin \frac{5x - x}{2} + 2 \cos 3x = 0$, $2 \sin 2x \cos 3x + 2 \cos 3x = 0, \quad 2 \cos 3x (\sin 2x + 1) = 0$.

## Hence

1) $\cos 3x = 0, \quad 3x = \frac{\pi}{2} + \pi k, \quad x_1 = \frac{\pi}{6} + \frac{\pi k}{3} = \frac{\pi}{6}(2k + 1), k \in \mathbb{Z}$;
2) $\sin 2x + 1 = 0, \quad \sin 2x = -1, \quad 2x = -\frac{\pi}{2} + 2 \pi n, \quad x_2 = -\frac{\pi}{4} + \pi n = \frac{\pi}{4}(4n - 1) \quad n \in \mathbb{Z}$.

Answer: $x_1 = \frac{\pi}{6}(2k + 1); x_2 = \frac{\pi}{4}(4n - 1), \quad k, n \in \mathbb{Z}$.