8.060. $\sin 2 x-\sin 3 x+\sin 8 x=\cos \left(7 x+\frac{3 \pi}{2}\right)$

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8.060. $\sin 2 x-\sin 3 x+\sin 8 x=\cos \left(7 x+\frac{3 \pi}{2}\right)$
## Solution.

From the condition

$$
\begin{aligned}
& (\sin 2 x - \sin 3 x) + (\sin 8 x - \sin 7 x) = 0, \Leftrightarrow 2 \sin \frac{2 x - 3 x}{2} \cos \frac{2 x + 3 x}{2} + \\
& + 2 \sin \frac{8 x - 7 x}{2} \cos \frac{8 x + 7 x}{2} = 0, -\sin \frac{x}{2} \cos \frac{5 x}{2} + \sin \frac{x}{2} \cos \frac{15 x}{2} = 0 \Leftrightarrow \\
& \Leftrightarrow -\sin \frac{x}{2} \left( \cos \frac{5 x}{2} - \cos \frac{15 x}{2} \right) = 0 \Leftrightarrow \\
& \Leftrightarrow \sin \frac{x}{2} \cdot 2 \sin \frac{\frac{5 x}{2} + \frac{15 x}{2}}{2} \sin \frac{\frac{15 x}{2} - \frac{5 x}{2}}{2} = 0 \Leftrightarrow \\
& \Leftrightarrow \sin \frac{x}{2} \sin 5 x \sin \frac{5}{2} x = 0
\end{aligned}
$$

Then:

1) $\sin \frac{x}{2} = 0, \frac{x}{2} = \pi n, x_1 = 2 \pi n, n \in \mathbb{Z}$;
2) $\sin 5 x = 0, 5 x = \pi k, x_2 = \frac{\pi k}{5}, k \in \mathbb{Z}$;
3) $\sin \frac{5}{2} x = 0, \frac{5}{2} x = \pi m, x_3 = \frac{2}{5} \pi m, m \in \mathbb{Z}; x_1$ and $x_3$ are included in $x_2$.

Answer: $x = \frac{\pi k}{5}, k \in \mathbb{Z}$.