8.276. $\operatorname{tg}^{4} x=36 \cos ^{2} 2 x$.

8.276. $\tan^{4} x=36 \cos ^{2} 2 x$.
## Solution.

Domain of definition: $\cos x \neq 0$.

From the condition, we have

$$
\begin{aligned}
& \frac{\left(\sin ^{2} x\right)^{2}}{\left(\cos ^{2} x\right)^{2}}-36 \cos ^{2} 2 x=0 \Leftrightarrow \frac{(1-\cos 2 x)^{2}}{(1+\cos 2 x)^{2}}-36 \cos ^{2} 2 x=0 \Leftrightarrow \\
& \Leftrightarrow(1-\cos 2 x)^{2}-36 \cos ^{2} 2 x(1+\cos 2 x)^{2}=0 \Leftrightarrow \\
& \Leftrightarrow(1-\cos 2 x)^{2}=36 \cos ^{2} 2 x(1+\cos 2 x)^{2}
\end{aligned}
$$

From this,

$1-\cos 2 x=-6 \cos 2 x(1+\cos 2 x)$

or

$$
1-\cos 2 x=6 \cos 2 x(1+\cos 2 x) 6 \cos ^{2} 2 x+5 \cos 2 x+1=0
$$

or

$6 \cos ^{2} 2 x+7 \cos 2 x-1=0$.

Solving these equations as quadratic equations in $\cos 2 x$, we get

$$
\begin{aligned}
& \cos 2 x=-\frac{1}{2}, 2 x_{1}= \pm \frac{2 \pi}{3}+2 \pi k, x_{1}= \pm \frac{\pi}{3}+\pi k, k \in Z \\
& \cos 2 x=-\frac{1}{3}, 2 x_{2}= \pm \arccos \left(-\frac{1}{3}\right)+2 \pi n, x_{2}= \pm \frac{1}{2} \arccos \left(-\frac{1}{3}\right)+\pi n, n \in Z
\end{aligned}
$$

$$
\begin{aligned}
& \cos 2 x=\frac{-7+\sqrt{73}}{12}, 2 x_{3}= \pm \arccos \frac{-7+\sqrt{73}}{12}+2 \pi l, \\
& x_{3}= \pm \frac{1}{2} \arccos \frac{-7+\sqrt{73}}{12}+\pi l, l \in Z ; \cos 2 x=\frac{-7-\sqrt{73}}{12}<-1, \varnothing
\end{aligned}
$$

Answer: $x_{1}= \pm \frac{\pi}{3}+\pi k, x_{2}= \pm \frac{1}{2} \arccos \left(-\frac{1}{3}\right)+\pi n$,

$$
x_{3}= \pm \frac{1}{2} \arccos \frac{-7+\sqrt{73}}{12}+\pi l, \text { where } k, n \text { and } l \in Z
$$