8.076. $\sin x - \sin 2x + \sin 5x + \sin 8x = 0$.
Solution.

We have

$$
\begin{aligned}
& (\sin x - \sin 2x) + (\sin x + \sin 8x) = 0 \Leftrightarrow 2 \sin \frac{x - 2x}{2} \cos \frac{x + 2x}{2} + \\
& + 2 \sin \frac{5x + 8x}{2} \cos \frac{5x - 8x}{2} = 0 \Leftrightarrow -2 \sin \frac{x}{2} \cos \frac{3x}{2} + 2 \sin \frac{13x}{2} \times \\
& \times \cos \frac{3x}{2} = 0 \Leftrightarrow -2 \cos \frac{3x}{2} \left( \sin \frac{x}{2} - \sin \frac{13x}{2} \right) = 0 \Leftrightarrow \cos \frac{3x}{2} \times \\
& \times 2 \sin \frac{\frac{x}{2} - \frac{13x}{2}}{2} \cos \frac{\frac{x}{2} + \frac{13x}{2}}{2} = 0 \Leftrightarrow \cos \frac{3x}{2} \sin 3x \cos \frac{7x}{2} = 0
\end{aligned}
$$

From this,

1) $\cos \frac{3x}{2} = 0, \frac{3x}{2} = \frac{\pi}{2} + \pi k, \quad x_1 = \frac{\pi}{3} + \frac{2\pi k}{3} = \frac{\pi}{3}(2k + 1), k \in \mathbb{Z}$;
2) $\sin 3x = 0, 3x = \pi n, x_2 = \frac{\pi n}{3}, n \in \mathbb{Z}$
3) $\cos \frac{7x}{2} = 0, \frac{7x}{2} = \frac{\pi}{2} + \pi l, x_3 = \frac{\pi}{7} + \frac{2\pi l}{7} = \frac{\pi}{7}(2l + 1), \quad l \in \mathbb{Z}$;

$x_1$ is included in $x_2$.

Answer: $\quad x_1 = \frac{\pi n}{3}; x_2 = \frac{\pi}{7}(2l + 1) \quad n, l \in \mathbb{Z}$.