$8.483 \operatorname{tg} x-\sin 2 x-\cos 2 x+2\left(2 \cos x-\frac{1}{\cos x}\right)=0$.

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

$8.483 \tan x - \sin 2x - \cos 2x + 2\left(2 \cos x - \frac{1}{\cos x}\right) = 0$.
Solution. $\frac{\sin x}{\cos x}-2 \sin x \cdot \cos x+2\left(2 \cos x-\frac{1}{\cos x}\right)-\cos 2 x=0 \Leftrightarrow$

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

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

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

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

## $8.4845 \sin 2 x-11(\sin x+\cos x)+7=0$.

Solution. $10 \sin x \cdot \cos x-11(\sin x+\cos x)+5+2=0 \Leftrightarrow$ $\Leftrightarrow 5(\sin x+\cos x)^{2}-11(\sin x+\cos x)+2=0 \Leftrightarrow\left[\begin{array}{l}\sin x+\cos x=2, \\ \sin x+\cos x=\frac{1}{5}\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left[\begin{array}{l}\varnothing \\ \cos \left(x-\frac{\pi}{4}\right)=\frac{1}{5 \sqrt{2}}\end{array} \Leftrightarrow x=\frac{\pi}{4} \pm \arccos \frac{\sqrt{2}}{10}+2 \pi k\right.$.

Answer: $x=\frac{\pi}{4} \pm \arccos \frac{\sqrt{2}}{10}+2 \pi k$.

## $8.485 \frac{1}{\sin 5 x}-\operatorname{ctg} x=\operatorname{tg} \frac{x}{2}$.

Solution. $\frac{1}{\sin 5 x}-\frac{\cos x}{\sin x}=\frac{1-\cos x}{\sin x} \Leftrightarrow \frac{1}{\sin 5 x}=\frac{1}{\sin x} \Leftrightarrow\left\{\begin{array}{l}\sin x \neq 0, \\ \sin 5 x-\sin x=0\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}\sin x \neq 0, \\ \sin 2 x \cdot \cos 3 x=0\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin x \neq 0, \\ {\left[\begin{array}{l}\cos x=0, \\ \cos 3 x=0 .\end{array}\right.}\end{array} \Leftrightarrow\right.\right.$

1) $\cos x=0 \Leftrightarrow x=\frac{\pi}{2}+\pi m$.
2) $\left\{\begin{array}{l}x \neq \pi l, \\ x=\frac{\pi}{6}+\frac{\pi k}{3}\end{array} \Leftrightarrow x=\frac{\pi}{6}+\frac{\pi k}{3}\right.$.

If $\frac{\pi}{2}+\pi m=\frac{\pi}{6}+\frac{\pi k}{3}$, then $k=3 m+1$. This means that the solutions $x=\frac{\pi}{2}+\pi m \quad$ are contained in the solutions $\quad x=\frac{\pi}{6}+\frac{\pi k}{3}$.

Answer: $x=\frac{\pi}{6}(2 k+1)$.