8.112. $\operatorname{ctg}^{3} x+\sin ^{-2} x-3 \operatorname{ctg} x-4=0$.

8.112. $\cot^{3} x+\csc ^{2} x-3 \cot x-4=0$.
## Solution.

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

## From the condition

$$
\frac{\cos ^{3} x}{\sin ^{3} x}+\frac{1}{\sin ^{2} x}-\frac{3 \cos x}{\sin x}-4=0 \Rightarrow \cos ^{3} x+\sin x-3 \cos x \sin ^{2} x-4 \sin ^{3} x=0
$$

$\cos ^{3} x+\sin x\left(\sin ^{2} x+\cos ^{2} x\right)-3 \cos x \sin ^{2} x-4 \sin ^{3} x=0, \cos ^{3} x+$ $+\sin x \cos ^{2} x+\sin ^{3} x-3 \cos x \sin ^{2} x-4 \sin ^{3} x=0, \quad\left(\cos ^{3} x+\sin x \cos ^{2} x\right)$ ) $-3 \cos x \sin ^{2} x-3 \sin ^{3} x=0, \cos ^{2} x(\cos x+\sin x)-3 \sin ^{2} x(\cos x+\sin x)=0$, $(\cos x+\sin x)\left(\cos ^{2} x-3 \sin ^{2} x\right)=0$.

## From this

1) $\cos x+\sin x=0$
2) $\cos ^{2} x-3 \sin ^{2} x=0 \Rightarrow \operatorname{ctg} x=-1, x_{1}=\frac{3 \pi}{4}+\pi n, n \in Z$;

$$
\operatorname{ctg} x= \pm \sqrt{3}, x_{2}= \pm \frac{\pi}{6}+\pi k, k \in Z
$$

Answer: $\quad x_{1}=\frac{3 \pi}{4}+\pi n ; x_{2}= \pm \frac{\pi}{6}+\pi k, n, k \in Z$.