8.098. $\sin z+\sin 2 z+\sin 3 z=\cos z+\cos 2 z+\cos 3 z$.

8.098. $\sin z + \sin 2z + \sin 3z = \cos z + \cos 2z + \cos 3z$.
Solution.

Rewrite the given equation as

$$
\begin{aligned}
& 2 \sin \frac{z+3 z}{2} \cos \frac{z-3 z}{2}+\sin 2 z=2 \cos \frac{z+3 z}{2} \cos \frac{z-3 z}{2}+\cos 2 z \Leftrightarrow \\
& \Leftrightarrow 2 \sin 2 z \cos z+\sin 2 z=2 \cos 2 z \cos z+\cos 2 z, \sin 2 z(2 \cos z+1)- \\
& -\cos 2 z(2 \cos z+1)=0,(2 \cos z+1)(\sin 2 z-\cos 2 z)=0
\end{aligned}
$$

From this:

1) $2 \cos z+1=0, \cos z=-\frac{1}{2}, \quad z_{1}= \pm \frac{2}{3} \pi+2 \pi k=\frac{2}{3} \pi(3 k \pm 1), k \in Z$;
2) $\sin 2 z-\cos 2 z=0 \Leftrightarrow \operatorname{tg} 2 z=1,2 z=\frac{\pi}{4}+\pi n, z_{2}=\frac{\pi}{8}+\frac{\pi n}{2}=\frac{\pi}{8}(4 n+1)$, $n \in Z$.

Answer: $\quad z_{1}=\frac{2}{3} \pi(3 k \pm 1) \quad z_{2}=\frac{\pi}{8}(4 n+1), \quad k, n \in Z$.