8.294. $\frac{4 \cos ^{2} t-1}{\sin t}=\operatorname{ctg} t(1+2 \cos 2 t)$. 

Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.

8.294. $\frac{4 \cos ^{2} t-1}{\sin t}=\operatorname{ctg} t(1+2 \cos 2 t)$. 

The above text in English, keeping the original text's line breaks and format, is:

8.294. $\frac{4 \cos ^{2} t-1}{\sin t}=\operatorname{ctg} t(1+2 \cos 2 t)$.
## Solution.

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

Write the equation in the form

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

$\Leftrightarrow 4 \cos ^{2} t-1=\cos t\left(4 \cos ^{2} t-1\right),\left(4 \cos ^{2} t-1\right)-\cos t\left(4 \cos ^{2} t-1\right)=0$, $\left(4 \cos ^{2} t-1\right)(1-\cos t)=0$.

## From this

1) $4 \cos ^{2} t-1=0, \cos ^{2} t=\frac{1}{4}, \cos t= \pm \frac{1}{2}, t_{1}= \pm \frac{\pi}{3}+\pi k=\frac{\pi}{3}(3 k \pm 1), k \in Z$;
2) $1-\cos t=0, \cos t=1$ - does not satisfy the domain of definition.

Answer: $x=\frac{\pi}{3}(3 k \pm 1), k \in Z$.