$8.413 \cos x \cdot \sqrt{\tan^{2} x-\sin ^{2} x}+\sin x \cdot \sqrt{\cot^{2} x-\cos ^{2} x}=2 \sin x$.
Solution. The original equation is equivalent to the following:

$\cos x \cdot \sqrt{\sin ^{2} x \cdot\left(\frac{1}{\cos ^{2} x}-1\right)}+\sin x \cdot \sqrt{\cos ^{2} x \cdot\left(\frac{1}{\sin ^{2} x}-1\right)}=2 \sin x \Leftrightarrow$
$\Leftrightarrow \frac{\cos x \cdot \sin ^{2} x}{|\cos x|}+\frac{\sin x \cdot \cos ^{2} x}{|\sin x|}=2 \sin x \Leftrightarrow$

1) $\left\{\begin{array}{l}\sin x>0, \\ \cos x>0, \\ \sin ^{2} x+\cos ^{2} x=2 \sin x\end{array} \Leftrightarrow\left\{\begin{array}{l}\cos x>0, \\ \sin x=\frac{1}{2}\end{array} \Leftrightarrow x=\frac{\pi}{6}+2 \pi k\right.\right.$
2) $\left\{\begin{array}{l}\sin x>0, \\ \cos x<0, \\ \sin ^{2} x+\cos ^{2} x=2 \sin x\end{array} \Leftrightarrow\left\{\begin{array}{l}\cos x<0, \\ \sin x=\frac{1}{2}\end{array} \Leftrightarrow x=\frac{5\pi}{6}+2 \pi k\right.\right.$
3) $\left\{\begin{array}{l}\sin x<0, \\ \cos x>0, \\ \sin ^{2} x-\cos ^{2} x=2 \sin x\end{array} \Leftrightarrow\left\{\begin{array}{l}\cos x>0, \\ \sin x=\frac{1-\sqrt{3}}{2}\end{array} \Leftrightarrow x=\arcsin \left(\frac{1-\sqrt{3}}{2}\right)+2 \pi k\right.\right.$.
4) $\left\{\begin{array}{l}\sin x>0, \\ \cos x<0, \\ \cos ^{2} x-\sin ^{2} x=2 \sin x\end{array} \Leftrightarrow\left\{\begin{array}{l}\cos x<0, \\ \sin x=\frac{\sqrt{3}-1}{2}\end{array} \Leftrightarrow x=\pi+\arcsin \left(\frac{1-\sqrt{3}}{2}\right)+2 \pi k\right.\right.$.

Answer: $x_{1}=\frac{\pi}{6}+\pi k, x_{2}=\arcsin \frac{1-\sqrt{3}}{2}+\pi k$.