8.235. $\operatorname{ctg}^{4} x=\cos ^{3} 2 x+1$.

8.235. $\cot^{4} x=\cos ^{3} 2 x+1$.
## Solution.

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

Since $1+\operatorname{ctg}^{2} \alpha=\frac{1}{\sin ^{2} a}$, the equation becomes $\left(\frac{1}{\sin ^{2} x}-1\right)^{2}=\cos ^{3} 2 x+1 \Leftrightarrow \frac{1}{\sin ^{4} x}-\frac{2}{\sin ^{2} x}+1=\cos ^{3} 2 x+1 \Leftrightarrow$ $\Leftrightarrow \frac{1-2 \sin ^{2} x}{\sin ^{4} x}=\cos ^{3} 2 x \Leftrightarrow \cos 2 x=\cos ^{3} 2 x \sin ^{4} x \Leftrightarrow$ $\Leftrightarrow \cos 2 x\left(1-\cos ^{2} 2 x \sin ^{4} x\right)=0 \Leftrightarrow$ $\Leftrightarrow \cos 2 x\left(1-\cos 2 x \sin ^{2} x\right)\left(1+\cos 2 x \sin ^{2} x\right)=0$.

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

$$
\begin{aligned}
& \Leftrightarrow\left(1-\sin ^{2} x\right) \sin ^{2} x-\sin ^{4} x+1=0, \sin ^{2} x-2 \sin ^{4} x+1=0 \\
& 2 \sin ^{4} x-\sin ^{2} x-1=0
\end{aligned}
$$

Solving the biquadratic equation, we get $\sin ^{2} x=1, \sin ^{2} x=-\frac{1}{2}, \varnothing$; $\sin x= \pm 1, x_{2}=\frac{\pi}{2}+\pi n, n \in Z$.
3) $1+\cos 2 x \sin ^{2} x=0, 2 \sin ^{4} x-\sin ^{2} x+1 \neq 0(D<0), \varnothing$.

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