$8.472 \frac{12}{\cos ^{2} x}+\frac{1}{3} \operatorname{ctg}^{2} x+10\left(2 \operatorname{tg} x+\frac{\operatorname{ctg} x}{3}\right)=1$.

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

$8.472 \frac{12}{\cos ^{2} x}+\frac{1}{3} \cot^{2} x+10\left(2 \tan x+\frac{\cot x}{3}\right)=1$.
Solution. $12\left(1+\operatorname{tg}^{2} x\right)+10\left(2 \operatorname{tg} x+\frac{\operatorname{ctg} x}{3}\right)+\frac{1}{3} \operatorname{ctg}^{2} x=1 \Leftrightarrow$.

$\Leftrightarrow 11+3\left(4 \operatorname{tg}^{2} x+\frac{1}{9} \operatorname{ctg}^{2} x\right)+10\left(2 \operatorname{tg} x+\frac{1}{3} \operatorname{ctg} x\right)=0$.

Let $y=2 \operatorname{tg} x+\frac{1}{3} \operatorname{ctg} x$. Then $11+3\left(y^{2}-\frac{4}{3}\right)+10 y=0 \Leftrightarrow$
$\Leftrightarrow 3 y^{2}+10 y+7=0 \Leftrightarrow\left[\begin{array}{l}y=-1, \\ y=-\frac{7}{3}\end{array}\right.$

1) $2 \operatorname{tg} x+\frac{1}{3 \operatorname{tg} x}=-1 \Rightarrow \varnothing$.
2) $2 \operatorname{tg} x+\frac{1}{3 \operatorname{tg} x}=-\frac{7}{3} \Leftrightarrow\left[\begin{array}{ll}\operatorname{tg} x=-1, \\ \operatorname{tg} x=-\frac{1}{6}\end{array} \Leftrightarrow \begin{array}{l}x_{1}=-\frac{\pi}{4}+\pi k, \\ x_{2}=-\operatorname{arctg} \frac{1}{6}+\pi k\end{array}\right.$

Answer: $\quad x_{1}=\frac{\pi}{4}(4 k-1), x_{2}=-\operatorname{arctg} \frac{1}{6}+\pi k$.