$5.56 \sin 2 z-\sin 6 z+2=0$.
5.56 Since $|\sin \alpha| \leqslant 1$, the equation $\sin 2 z-\sin 6 z=-2$ is possible only when $\sin 2 z=-1, \sin 6 z=1$. If $\sin 2 z=-1$, then

$2 z=-\frac{\pi}{2}+2 \pi k$, from which $z=-\frac{\pi}{4}+\pi k$. For these values of $z$

the equation $\sin 6 z=1$ is satisfied.

Answer: $\quad z=\frac{\pi}{4}(4 k-1), k \in Z$.