8.043. $\sin 3z - \cos 3z = \sqrt{\frac{3}{2}}$.
Solution.

From the condition

$$
\begin{aligned}
& \sin 3 z \cdot \frac{\sqrt{2}}{2}-\cos 3 z \cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{3}}{2} \Leftrightarrow \sin 3 z \cos 45^{\circ}-\cos 3 z \sin 45^{\circ}=\frac{\sqrt{3}}{2} \Leftrightarrow \\
& \Leftrightarrow \sin \left(3 z-45^{\circ}\right)=\frac{\sqrt{3}}{2}
\end{aligned}
$$

from which $3 z-45^{\circ}=60^{\circ}+360^{\circ} k$ or $3 z-45^{\circ}=120^{\circ}+360^{\circ} k$. Hence, $z_{1}=35^{\circ}+120^{\circ} k, z_{2}=55^{\circ}+120^{\circ} k, k \in Z$.

Answer: $z_{1}=35^{\circ}+120^{\circ} k, z_{2}=55^{\circ}+120^{\circ} k, k \in Z$.