8.005. $\sin z \sin \left(60^{\circ}-z\right) \sin \left(60^{\circ}+z\right)=\frac{1}{8}$.
## Solution.

From the condition

$$
\begin{aligned}
& \sin z\left(2 \sin \left(60^{\circ}-z\right) \sin \left(60^{\circ}+z\right)\right)=\frac{1}{4}, \Leftrightarrow \sin z\left(\cos 2 z-\cos 120^{\circ}\right)=\frac{1}{4} \Leftrightarrow \\
& \Leftrightarrow 2 \sin z \cos 2 z+\sin z=\frac{1}{2}, \Leftrightarrow-\sin z+\sin 3 z+\sin z=\frac{1}{2}, \quad \sin 3 z=\frac{1}{2}
\end{aligned}
$$

Then $3 z=(-1)^{k} \cdot 30^{\circ}+180^{\circ} k, k \in Z, \quad z=(-1)^{k} \cdot 10^{\circ}+60^{\circ} k, k \in Z$.

Answer: $\quad z=(-1)^{k} \cdot 10^{\circ}+60^{\circ} k, k \in Z$.