8.244. $\operatorname{ctg}^{4} x=\cos ^{2} 2 x-1$.

8.244. $\cot^{4} x=\cos ^{2} 2 x-1$.
## Solution.

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

From the condition we have

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

From this, 1) $1+\cos 2 x=0$, 2) $\frac{1+\cos 2 x}{(1-\cos 2 x)^{2}}+(1-\cos 2 x)=0$.

From the first equation, we find $\cos 2 x=-1, 2 x=\pi+2 \pi k, x_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0528.jpg?height=69&width=896&top_left_y=1237&top_left_x=10)

$1+\cos 2 x+(1-\cos 2 x)^{3}=0, \varnothing$.

Answer: $x=\frac{\pi}{2}(2 k+1), k \in Z$