8.206. $\operatorname{tg}^{4} x+\operatorname{ctg}^{4} x=\frac{82}{9}(\operatorname{tg} x \operatorname{tg} 2 x+1) \cos 2 x$.
## Solution.

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

Write the equation in the form:

$$
\begin{aligned}
& \left((\operatorname{tg} x+\operatorname{ctg} x)^{2}-2 \operatorname{tg} x \operatorname{tg} x\right)^{2}-2 \operatorname{tg}^{2} x \operatorname{tg}^{2} x=\frac{82}{9} \cdot\left(\frac{\sin x \sin 2 x}{\cos x \cos 2 x}+1\right) \cos 2 x \Leftrightarrow \\
& \Leftrightarrow\left(\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)^{2}-2\right)^{2}-2=\frac{82}{9} \cdot\left(\frac{\sin x \sin 2 x+\cos x \cos 2 x}{\cos x \cos 2 x}\right) \cos 2 x
\end{aligned}
$$

Applying the formula $\cos \alpha \cos \beta+\sin \alpha \sin \beta=\cos (\alpha-\beta)$, we have

$\left(\left(\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right)^{2}-2\right)^{2}-2=\frac{82}{9} \cdot \frac{\cos x \cos 2 x}{\cos x \cos 2 x} \Leftrightarrow$

$\Leftrightarrow\left(\frac{1}{\sin ^{2} x \cos ^{2} x}-2\right)^{2}=\frac{100}{9}$.

From here 1) $\frac{1}{\sin ^{2} x \cos ^{2} x}-2= \pm \frac{10}{3} \Leftrightarrow \frac{4}{\sin ^{2} 2 x}-2= \pm \frac{10}{3}$. Then $\sin ^{2} 2 x=-3, \varnothing$; or $\sin ^{2} 2 x=\frac{3}{4}, \sin 2 x= \pm \frac{\sqrt{3}}{2}, 2 x= \pm \frac{\pi}{3}+\pi k$, $x= \pm \frac{\pi}{6}+\frac{\pi k}{2}=\frac{\pi}{6}(3 k \pm 1), k \in Z$.

Answer: $x=\frac{\pi}{6}(3 k \pm 1), k \in Z$