8.372. $\cos 3 x=m \cos x$.

8.372. $\cos 3 x=m \cos x$.
## Solution.

From the formula $\cos 3 \alpha=4 \cos ^{3} \alpha-3 \cos \alpha$, we have $4 \cos ^{3} x-3 \cos x-m \cos x=0, \cos x\left(4 \cos ^{2} x-3-m\right)=0$.

## Hence

1) $\cos x=0, x_{1}=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2 k+1), k \in Z$;
2) $4 \cos ^{2} x-3-m=0 \Leftrightarrow 2(1+\cos 2 x)=m+3, \cos 2 x=\frac{m+1}{2}$,

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0610.jpg?height=200&width=973&top_left_y=185&top_left_x=65)

Answer: $x_{1}=\frac{\pi}{2}(2 k+1)$, where $m \in R ; x_{2}= \pm \frac{1}{2} \arccos \frac{m+1}{2}+\pi n$, where $m \in[-3 ; 1] ; k$ and $n \in Z$.