8.262. $\operatorname{tg} z \operatorname{tg}\left(z+60^{\circ}\right) \operatorname{tg}\left(z+120^{\circ}\right)=\sqrt{3}$.
Solution.

$$
\text { Domain of definition: }\left\{\begin{array}{l}
\cos z \neq 0 \\
\cos \left(z+60^{\circ}\right) \neq 0 \\
\cos \left(z+120^{\circ}\right) \neq 0
\end{array}\right.
$$

Write the equation in the form

$\frac{\sin z \sin \left(z+60^{\circ}\right) \sin \left(z+120^{\circ}\right)}{\cos z \cos \left(z+60^{\circ}\right) \cos \left(z+120^{\circ}\right)}-\sqrt{3}=0 \Leftrightarrow$

$\Leftrightarrow \sin z\left(2 \sin \left(z+60^{\circ}\right) \sin \left(z+120^{\circ}\right)\right)-\sqrt{3} \cdot \cos z\left(2 \cos \left(z+60^{\circ}\right) \cos \left(z+120^{\circ}\right)\right)=0$.

Using the formulas

$\sin \alpha \sin \beta=\frac{1}{2}(\cos (\alpha-\beta)-\cos (\alpha+\beta))$

and

$\cos \alpha \cos \beta=\frac{1}{2}(\cos (\alpha-\beta)+\cos (\alpha+\beta))$,

we have:

$$
\begin{aligned}
& \sin z\left(\cos \left(-60^{\circ}\right)-\cos \left(180^{\circ}+2 z\right)\right)-\sqrt{3} \cos z \cdot\left(\cos \left(-60^{\circ}\right)+\cos \left(180^{\circ}+2 z\right)\right)=0, \\
& \sin z\left(\frac{1}{2}+\cos 2 z\right)-\sqrt{3} \cos z\left(\frac{1}{2}-\cos 2 z\right)=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\sin z}{2}+\sin z \cos 2 z-\frac{\sqrt{3}}{2} \cos z+\sqrt{3} \cos z \cos 2 z=0 \Leftrightarrow \\
& \Leftrightarrow \sin z+2 \sin z \cos 2 z-\sqrt{3} \cos z+\sqrt{3}(2 \cos z \cos 2 z)=0 \Leftrightarrow \\
& \Leftrightarrow \sin z-\sin z+\sin 3 z-\sqrt{3} \cos z+\sqrt{3} \cos z+\sqrt{3} \cos 3 z=0 \Leftrightarrow \\
& \Leftrightarrow \sin 3 z+\sqrt{3} \cos 3 z=0 \Leftrightarrow \operatorname{tg} 3 z=-\sqrt{3}, 3 z=-60^{\circ}+180^{\circ} k \\
& z=-20^{\circ}+60^{\circ} k
\end{aligned}
$$

$$
\text { Answer: } z=-20^{\circ}+60^{\circ} k, k \in Z \text {. }
$$