$8.408 \operatorname{tg} x-\sin 2 x-\cos 2 x \cdot\left(1-\frac{2}{\cos x}\right)=0$.

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

$8.408 \tan x - \sin 2x - \cos 2x \cdot \left(1 - \frac{2}{\cos x}\right) = 0$.
Solution. $\frac{\sin x}{\cos x}-\sin 2 x-\cos 2 x \cdot \frac{\cos x-2}{\cos x}=0 \Leftrightarrow$ 142
$\Leftrightarrow\left\{\begin{array}{l}\sin x-\sin 2 x \cdot \cos x-\cos 2 x \cdot(\cos x-2)=0, \\ \cos x \neq 0\end{array} \Leftrightarrow\right.$

$\Leftrightarrow\left\{\begin{array}{l}\sin x \cdot\left(1-2 \cos ^{2} x\right)+\left(1-2 \cos ^{2} x\right) \cdot(\cos x-2)=0, \\ \cos x \neq 0\end{array} \Leftrightarrow\right.$

$\Leftrightarrow\left(1-2 \cos ^{2} x\right) \cdot(\sin x+\cos x-2)=0 \Leftrightarrow$

$\Leftrightarrow\left[\begin{array}{l}\cos 2 x=0, \\ \sin x+\cos x=2\end{array} \Leftrightarrow\left[\begin{array}{l}\cos 2 x=0, \\ \{\sin x=1, \\ \cos x=1\end{array} \Leftrightarrow \cos 2 x=0^{\circ} \Leftrightarrow x=\frac{\pi}{4}(2 k+1)\right.\right.$

Answer: $x=\frac{\pi}{4}(2 k+1)$.