$8.457 \cos 6 x+\sin \frac{5 x}{2}=2$.

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

$8.457 \cos 6 x+\sin \frac{5 x}{2}=2$.
Solution. The equation is equivalent to the system:
$\left\{\begin{array}{l}\cos 6 x=1, \\ \sin \frac{5 x}{2}=1\end{array} \Leftrightarrow\left\{\begin{array}{c}x=\frac{\pi l}{3}, l \in Z, \\ x=\frac{\pi}{5}+\frac{4 \pi n}{5}, n \in Z\end{array} \Leftrightarrow \frac{\pi l}{3}=\frac{\pi}{5}+\frac{4 \pi n}{5} \Rightarrow l=3+12 k, k \in Z\right.\right.$

The solution of the equation (4)) $\Rightarrow x=\frac{\pi(3+12 k)}{3}=\pi+4 \pi k$.

Answer: $x=\pi(1+4 k)$.

## $8.458 \sqrt{3}|\cos x|=1+\operatorname{ctg} x$.

## Solution.

$\sqrt{3}|\cos x|=\frac{\sin x+\cos x}{\sin x} \Leftrightarrow 2 \sin x|\cos x|=\frac{2}{\sqrt{3}}(\sin x+\cos x)$

1) $\left\{\begin{array}{l}\cos x \geq 0, \\ 1+2 \sin x \cdot \cos x=\frac{2}{\sqrt{3}}(\sin x+\cos x)+1\end{array} \Leftrightarrow\right.$

$\Leftrightarrow\left\{\begin{array}{l}\cos x \geq 0, \\ (\sin x+\cos x)^{2}-\frac{2}{\sqrt{3}}(\sin x+\cos x)-1=0\end{array} \Leftrightarrow\left\{\begin{array}{l}\cos x \geq 0, \\ \cos x+\sin x=\frac{1 \pm 2}{\sqrt{3}}\end{array} \Leftrightarrow\right.\right.$

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

If $l=2 k$, then $x=-\left(\frac{\pi}{4}+\arcsin \frac{1}{\sqrt{6}}\right)+2 \pi k$ and $x$ is in the fourth quadrant, that is, $\cos x \geq 0$. If $l=2 k+1$, then $x=\frac{3 \pi}{4}+\arcsin \frac{1}{\sqrt{6}}+2 \pi k$, that is, $\cos x<0$. If $l=2 k+1$, then $x=\frac{3 \pi}{4}-\arcsin \frac{1}{\sqrt{6}}+2 \pi k$ and $x$ is in the second quadrant.

Thus, $x=\frac{3 \pi}{4}-\arcsin \frac{1}{\sqrt{6}}+2 \pi k$.

Answer: $x=\frac{\pi}{4}(4 k-1)-\arcsin \frac{1}{\sqrt{6}}$.