8.234. $\frac{1}{2}\left(\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x\right)=1+\frac{2}{\sqrt{3}} \operatorname{ctg} 2 x$.

8.234. $\frac{1}{2}\left(\tan^{2} x+\cot^{2} x\right)=1+\frac{2}{\sqrt{3}} \cot 2 x$.
## Solution.

Domain of definition: $\sin 2 x \neq 0$.

## Write the equation in the form

$1+1+\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x=4+\frac{4 \operatorname{ctg} 2 x}{\sqrt{3}} \Leftrightarrow$

$\Leftrightarrow \frac{1}{\cos ^{2} x}+\frac{1}{\sin ^{2} x}=\frac{4(\sqrt{3}+\operatorname{ctg} 2 x)}{\sqrt{3}} \Leftrightarrow \frac{4}{\sin ^{2} 2 x}=\frac{4(\sqrt{3}+\operatorname{ctg} 2 x)}{\sqrt{3}} \Leftrightarrow$

$\Leftrightarrow \sqrt{3} \sin ^{2} 2 x+\sin 2 x \cos 2 x-\sqrt{3}=0 \Leftrightarrow$

$\Leftrightarrow-\sqrt{3}\left(1-\sin ^{2} 2 x\right)+\sin 2 x \cos 2 x=0 \Leftrightarrow$

$\Leftrightarrow \sqrt{3} \cos ^{2} 2 x-\sin 2 x \cos 2 x=0 \Leftrightarrow \cos 2 x(\sqrt{3} \cos 2 x-\sin 2 x)=0$.

From this,

1) $\cos 2 x=0, 2 x=\frac{\pi}{2}+\pi k, x_{1}=\frac{\pi}{4}+\frac{\pi k}{2}=\frac{\pi}{4}(2 k+1), k \in Z$;
2) $\sqrt{3} \cos 2 x-\sin 2 x=0 \Leftrightarrow \operatorname{ctg} 2 x=\frac{1}{\sqrt{3}}, 2 x=\frac{\pi}{3}+\pi n$,

$$
x_{2}=\frac{\pi}{6}+\frac{\pi n}{2}=\frac{\pi}{6}(3 n+1), n \in Z
$$

Answer: $x_{1}=\frac{\pi}{4}(2 k+1) ; x_{2}=\frac{\pi}{6}(3 n+1)$, where $k$ and $n \in Z$.