8.308. $\cos 3z - \cos^3 z + \frac{3}{4} \sin 2z = 0$.
Solution.

Since $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$, we obtain

$$
\begin{aligned}
& 4\left(4 \cos ^{3} z-3 \cos z\right)-4 \cos ^{3} z+6 \sin z \cos z=0 \\
& 12 \cos ^{3} z-12 \cos z+6 \sin z \cos z=0,6 \cos z\left(2 \cos ^{2} z-2+\sin z\right)=0
\end{aligned}
$$

$6 \cos z\left(2\left(1-\sin ^{2} z\right)-2+\sin z\right)=0,6 \cos z\left(-2 \sin ^{2} z+\sin z\right)=0$,

$6 \cos z \sin z(2 \sin z-1)=0$.

From this,

1) $\cos z=0, z_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z$;
2) $\sin z=0, z_{2}=\pi n, n \in Z$
3) $2 \sin z-1=0, \sin z=\frac{1}{2}, z_{3}=(-1)^{k} \frac{\pi}{6}+\pi l, l \in Z$.

Answer: $z_{1}=\frac{\pi}{2}(2 k+1) ; z_{2}=\pi n ; z_{3}=(-1)^{k} \frac{\pi}{6}+\pi l$, where $k, n$ and $l \in Z$.