8.217. $\left(\cos 2 x+(\cos x+\sin x)^{2}\right)(\tan x+\cot x)=0$.
Solution.

Domain of definition: $\left\{\begin{array}{l}\cos x \neq 0, \\ \sin x \neq 0 .\end{array}\right.$

The equation is equivalent to two equations: $\cos 2 x+(\cos x+\sin x)^{2}=0$ or $\operatorname{tg} x+\operatorname{ctg} x=0$. Let's write the first equation as

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

Since $\cos x \neq 0$, then

$$
\cos x+\sin x=0 \Leftrightarrow \operatorname{tg} x=-1, x=-\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n-1), n \in Z
$$

Let's write the second equation as

$$
\operatorname{tg} x+\frac{1}{\operatorname{ctg} x}=0 \Leftrightarrow \frac{\operatorname{tg}^{2} x+1}{\operatorname{tg} x}=0 \Leftrightarrow \operatorname{tg}^{2} x=-1, \varnothing
$$

Answer: $x=\frac{\pi}{4}(4 n-1), n \in Z$