8.068. $\operatorname{tg}\left(\frac{\pi}{2}+x\right)-\operatorname{ctg}^{2} x-\sin ^{-2} x(1+\cos 2 x)=0$.

8.068. $\tan\left(\frac{\pi}{2}+x\right)-\cot^{2} x-\csc ^{2} x(1+\cos 2 x)=0$.
## Solution.

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

From the condition

$$
\begin{aligned}
& -\operatorname{ctg} x-\operatorname{ctg}^{2} x+\frac{1+2 \cos ^{2} x-1}{\sin ^{2} x}=0,-\operatorname{ctg} x-\operatorname{ctg}^{2} x+2 \operatorname{ctg}^{2} x=0, \\
& \operatorname{ctg}^{2} x-\operatorname{ctg} x=0, \operatorname{ctg} x(\operatorname{ctg} x-1)=0
\end{aligned}
$$

From this:

1) $\operatorname{ctg} x=0, \quad x_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), \quad k \in Z$;
2) $\operatorname{ctg} x-1=0, \operatorname{ctg} x=1, \quad x_{2}=\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n+1), \quad n \in Z$.

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