8.114. $1+\sin x-\cos 5 x-\sin 7 x=2 \cos ^{2} \frac{3}{2} x$.

8.114. $1+\sin x-\cos 5x-\sin 7x=2 \cos^2 \frac{3}{2}x$.
## Solution.

Let's rewrite the given equation as

$1+\sin x-\cos 5 x-\sin 7 x=1+\cos 3 x \Leftrightarrow (\sin x-\sin 7 x)-(\cos 5 x+\cos 3 x)=0 \Leftrightarrow$ $\Leftrightarrow -2 \sin 3 x \cos 4 x - 2 \cos 4 x \cos x = 0, \quad -2 \cos 4 x (\sin 3 x + \cos x) = 0$.

From this, we have:

1) $\cos 4 x = 0, 4 x = \frac{\pi}{2} + \pi k, x_{1} = \frac{\pi}{8} + \frac{\pi k}{4} = \frac{\pi}{8}(2 k + 1), k \in Z$;
2) $\sin 3 x + \cos x = 0 \Leftrightarrow \sin 3 x + \sin \left(\frac{\pi}{2} - x\right) = 0, 2 \sin \frac{3 x + \frac{\pi}{2} - x}{2} \times$

$$
\times \cos \frac{3 x - \frac{\pi}{2} + x}{2} = 0, \quad \sin \left(x + \frac{\pi}{4}\right) \cos \left(2 x - \frac{\pi}{4}\right) = 0
$$

We have

$$
\sin \left(x + \frac{\pi}{4}\right) = 0, \quad x + \frac{\pi}{4} = \pi n, \quad x_{2} = -\frac{\pi}{4} + \pi n = \frac{\pi}{4}(4 n - 1) \quad n \in Z
$$

$$
\cos \left(2 x - \frac{\pi}{4}\right) = 0, \quad 2 x - \frac{\pi}{4} = \frac{\pi}{2} + \pi l, \quad x_{3} = \frac{3}{8} \pi + \frac{\pi l}{2}, \quad l \in Z
$$

$x_{3}$ is included in $x_{1}$.

Answer: $\quad x_{1} = \frac{\pi}{8}(2 k + 1), x_{2} = \frac{\pi}{4}(4 n - 1) \quad k, n \in Z$.