8.187. $\sin ^{2} x \tan x+\cos ^{2} x \cot x+2 \sin x \cos x=\frac{4 \sqrt{3}}{3}$.
Solution.

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

Write the equation in the form:

$$
\begin{aligned}
& \frac{\sin ^{2} x \sin x}{\cos x}+\frac{\cos ^{2} x \cos x}{\sin x}+2 \sin x \cos x-\frac{4 \sqrt{3}}{3}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\sin ^{4} x+\cos ^{4} x}{\sin x \cos x}+2 \sin x \cos x-\frac{4 \sqrt{3}}{3}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-2 \sin ^{2} x \cos ^{2} x}{\sin x \cos x}+2 \sin x \cos x-\frac{4 \sqrt{3}}{3}=0
\end{aligned}
$$

or

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

$-2 \sin x \cos x+2 \sin x \cos x-\frac{4 \sqrt{3}}{3}=0 \Leftrightarrow \frac{1}{2 \sin x \cos x}=\frac{2 \sqrt{3}}{2} \Leftrightarrow$ $\Leftrightarrow \sin 2 x=\frac{\sqrt{3}}{2} \Rightarrow 2 x=(-1)^{k}, \frac{\pi}{3}+\pi k, x=(-1)^{k} \frac{\pi}{6}+\frac{\pi}{2}, k \in Z$.

Answer: $x=(-1)^{k} \frac{\pi}{6}+\frac{\pi}{2}, k \in Z$.