8.074. $1-\cos (\pi+x)-\sin \frac{3 \pi+x}{2}=0$.
Solution.

We have

$1+\cos x+\cos \frac{x}{2}=0 \Leftrightarrow 1+2 \cos ^{2} \frac{x}{2}-1+\cos \frac{x}{2}=0$,

$2 \cos ^{2} \frac{x}{2}+\cos \frac{x}{2}=0, \quad \cos \frac{x}{2}\left(2 \cos \frac{x}{2}+1\right)=0$.

From this:

1) $\cos \frac{x}{2}=0, \quad \frac{x}{2}=\frac{\pi}{2}+\pi k, \quad x_{1}=\pi+2 \pi k=\pi(2 k+1), \quad k \in Z$;
2) $2 \cos \frac{x}{2}+1=0, \cos \frac{x}{2}=-\frac{1}{2}, \frac{x}{2}= \pm \frac{2}{3} \pi+2 \pi n, \quad x_{2}= \pm \frac{4}{3} \pi+4 \pi n$, $n \in Z$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-453.jpg?height=104&width=792&top_left_y=970&top_left_x=88)