8.322. $3 \tan 3 x-4 \tan 2 x=\tan^{2} 2 x \tan 3 x$.
## Solution.

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

## Write the equation in the form

$$
\begin{aligned}
& 3 \operatorname{tg} 3 x-3 \operatorname{tg} 2 x=\operatorname{tg}^{2} 2 x \operatorname{tg} 3 x+\operatorname{tg} 2 x \\
& 3 \cdot\left(\frac{\sin 3 x}{\cos 3 x}-\frac{\sin 2 x}{\cos 2 x}\right)=\operatorname{tg} 2 x \cdot\left(\frac{\sin 2 x \sin 3 x}{\cos 2 x \cos 3 x}+1\right) \Leftrightarrow \\
& \Leftrightarrow 3 \cdot\left(\frac{\sin 3 x \cos 2 x-\cos 3 x \sin 2 x}{\cos 3 x \cos 2 x}\right)-\frac{\sin 2 x}{\cos 2 x} \times \\
& \times\left(\frac{\sin 2 x \sin 3 x+\cos 2 x \cos 3 x}{\cos 2 x \cos 3 x}\right)=0 \Leftrightarrow \\
& \Leftrightarrow \frac{3 \sin x}{\cos 3 x \cos 2 x}-\frac{\sin 2 x \cos x}{\cos ^{2} 2 x \cos 3 x}=0 \Leftrightarrow \\
& \Leftrightarrow 3 \sin x \cos 2 x-\sin 2 x \cos x=0,3 \sin x\left(2 \cos ^{2} x-1\right)-2 \sin x \cos ^{2} x=0, \\
& \sin x\left(4 \cos ^{2} x-3\right)=0
\end{aligned}
$$

## From here

1) $\sin x=0, x_{1}=\pi k, k \in Z$;
2) $4 \cos ^{2} x-3=0, \cos x= \pm \frac{\sqrt{3}}{2}, x_{2}= \pm \frac{\pi}{6}+\pi n, n \in Z ; x_{2}$ does not satisfy the domain of definition.

Answer: $x=\pi k, k \in Z$.