8.285. $\frac{1}{\sin ^{2} 2 x}+\tan x-\cot x-4=0$.
Solution.

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

Rewrite the equation as

$\frac{1}{\sin ^{2} 2 x}+\left(\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right)-4=0 \Leftrightarrow \frac{1}{\sin ^{2} 2 x}+\frac{\sin ^{2} x-\cos ^{2} x}{\sin x \cos x}-4=0 \Leftrightarrow$ $\frac{1}{\sin ^{2} 2 x}-\frac{2 \cos 2 x}{\sin 2 x}-4=0 \Leftrightarrow 1-2 \sin 2 x \cos 2 x-4 \sin ^{2} 2 x=0$,
$\cos ^{2} 2 x+\sin ^{2} 2 x-2 \sin 2 x \cos 2 x-4 \sin ^{2} 2 x=0 \Leftrightarrow$

$\Leftrightarrow \cos ^{2} 2 x-2 \sin 2 x \cos 2 x-3 \sin ^{2} 2 x=0 \Leftrightarrow \operatorname{ctg}^{2} 2 x-2 \operatorname{ctg} 2 x-3=0$.

Solve this equation as a quadratic equation in terms of $\operatorname{ctg} 2 x$:

$\operatorname{ctg} 2 x=-1,2 x=\frac{3 \pi}{4}+\pi k, x_{1}=\frac{3 \pi}{8}+\frac{\pi k}{2}, k \in Z$

$\operatorname{ctg} 2 x=3,2 x=\operatorname{arcctg} 3+\pi n, x_{2}=\frac{1}{2} \operatorname{arcctg} 3+\frac{\pi n}{2}, n \in Z$.

Answer: $x_{1}=\frac{3 \pi}{8}+\frac{\pi k}{2} ; x_{2}=\frac{1}{2} \operatorname{arcctg} 3+\frac{\pi n}{2}$, where $k$ and $n \in Z$.