8.351. $3 \sqrt{3} \tan x \sin x - \cot x \cos x + 9 \sin x - 3 \sqrt{3} \cos x = 0$.
## Solution.

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

Rewrite the equation as

$$
\begin{aligned}
& \frac{3 \sqrt{3} \sin ^{2} x}{\cos x}-\frac{\cos ^{2} x}{\sin x}+9 \sin x-3 \sqrt{3} \cos x=0 \Leftrightarrow \\
& \Leftrightarrow 3 \sqrt{3} \sin ^{3} x+9 \sin ^{2} x \cos x-3 \sqrt{3} \cos ^{2} x \sin x-\cos ^{3} x=0 \Leftrightarrow \\
& \Leftrightarrow 3 \sqrt{3} \operatorname{tg}^{3} x+9 \operatorname{tg}^{2} x-3 \sqrt{3} \operatorname{tg} x-1=0 \Leftrightarrow \\
& \Leftrightarrow\left(3 \sqrt{3} \operatorname{tg}^{3} x-1\right)+\left(9 \operatorname{tg}^{2} x-3 \sqrt{3} \operatorname{tg} x\right)=0 \Leftrightarrow \\
& \Leftrightarrow(\sqrt{3} \operatorname{tg} x-1)\left(3 \operatorname{tg}^{2} x+\sqrt{3} \operatorname{tg} x+1\right)+3 \sqrt{3}(\sqrt{3} \operatorname{tg} x-1) \operatorname{tg} x=0 \Leftrightarrow \\
& \Leftrightarrow(\sqrt{3} \operatorname{tg} x-1)\left(3 \operatorname{tg}^{2} x+4 \sqrt{3} \operatorname{tg} x+1\right)=0 .
\end{aligned}
$$

From this,

1) $\sqrt{3} \operatorname{tg} x-1=0, \operatorname{tg} x=\frac{1}{\sqrt{3}}, x_{1}=\frac{\pi}{6}+\pi k=\frac{\pi}{6}(6 k+1), k \in Z$;
2) $3 \operatorname{tg}^{2} x+4 \sqrt{3} \operatorname{tg} x+1=0$. Solving this equation as a quadratic in $\operatorname{tg} x$, we find

$$
\operatorname{tg} x=\frac{-2 \sqrt{3}-3}{3}, x_{2}=\operatorname{arctg} \frac{-2 \sqrt{3}-3}{3}+\pi n, n \in Z
$$

or

$$
\operatorname{tg} x=\frac{-2 \sqrt{3}+3}{3}, x_{3}=\operatorname{arctg} \frac{-2 \sqrt{3}+3}{3}+\pi l, l \in Z
$$

Answer: $x_{1}=\frac{\pi}{6}(6 k+1), x_{2}=\operatorname{arctg} \frac{-2 \sqrt{3}-3}{3}+\pi n$,

$$
x_{3}=\operatorname{arctg} \frac{-2 \sqrt{3}+3}{3}+\pi l, \text { where } k, n \text { and } l \in Z
$$