8.106. $\cos \left(3 x-30^{\circ}\right)-\sin \left(3 x-30^{\circ}\right) \tan 30^{\circ}=\frac{1}{2 \cos 210^{\circ}}$.
## Solution.

From the condition

$$
\begin{aligned}
& \cos \left(3 x-30^{\circ}\right)-\frac{\sin \left(3 x-30^{\circ}\right) \sin 30^{\circ}}{\cos 30^{\circ}}=\frac{1}{2 \cos \left(180^{\circ}+30^{\circ}\right)} \Leftrightarrow \\
& \frac{\cos \left(3 x-30^{\circ}\right) \cos 30^{\circ}-\sin \left(3 x-30^{\circ}\right) \sin 30^{\circ}}{\cos 30^{\circ}}-\frac{1}{2 \cos \left(180^{\circ}+30^{\circ}\right)}=0 \Leftrightarrow \\
& \frac{\cos \left(3 x-30^{\circ}+30^{\circ}\right)}{\cos 30^{\circ}}+\frac{1}{2 \cos 30^{\circ}}=0 \Leftrightarrow 2 \cos 3 x+1=0
\end{aligned}
$$

from which $\cos 3 x=-\frac{1}{2}, 3 x= \pm \frac{2}{3} \pi+2 \pi k, \quad x= \pm \frac{2 \pi}{9}+\frac{2}{3} \pi k, \quad k \in Z$.

Answer: $\quad x= \pm \frac{2 \pi}{9}+\frac{2}{3} \pi k, k \in Z$.