8.026. $\cos ^{-1} x+\operatorname{ctg} 3 x=\operatorname{ctg} \frac{3 x}{2}$.

8.026. $\cos ^{-1} x+\cot 3 x=\cot \frac{3 x}{2}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin 3 x \neq 0, \\ \sin \frac{3 x}{2} \neq 0\end{array}\right.$

## From the condition

$$
\begin{aligned}
& \frac{1}{\cos x}+\left(\operatorname{ctg} 3 x-\operatorname{ctg} \frac{3 x}{2}\right)=0 \Leftrightarrow \frac{1}{\cos x}-\frac{\sin \left(3 x-\frac{3}{2} x\right)}{\sin 3 x \sin \frac{3}{2} x}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{1}{\cos x}-\frac{\sin \frac{3}{2} x}{\sin 3 x \sin \frac{3}{2} x}=0, \frac{1}{\cos x}-\frac{1}{\sin 3 x}=0, \frac{\sin 3 x-\cos x}{\cos x \sin 3 x}=0
\end{aligned}
$$

$\sin 3 x-\cos x=0, \Leftrightarrow \sin 3 x-\sin \left(\frac{\pi}{2}-x\right)=0 \Leftrightarrow$

$\Leftrightarrow 2 \cos \frac{3 x+\frac{\pi}{2}-x}{2} \cdot \sin \frac{3 x-\frac{\pi}{2}+x}{2}=0$,

from which: 1) $\left.\cos \left(x+\frac{\pi}{4}\right)=0 ; 2\right) \sin \left(2 x-\frac{\pi}{4}\right)=0$.

Then:

1) $x+\frac{\pi}{4}=\frac{\pi}{2}+\pi k, \quad x_{1}=\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k+1), k \in Z$;
2) $2 x-\frac{\pi}{4}=\pi n, \quad x_{2}=\frac{\pi}{8}+\frac{\pi n}{2}=\frac{\pi}{8}(4 n+1), \quad n \in Z$.

Answer: $\quad x_{1}=\frac{\pi}{4}(4 k+1) ; x_{2}=\frac{\pi}{8}(4 n+1) \quad k, n \in Z$.