8.324. $4 \cos x=\sqrt{3} \operatorname{ctg} x+1$.

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

8.324. $4 \cos x=\sqrt{3} \cot x+1$.
## Solution.

Domain of definition: $\sin x \neq 0$.

Let's write the equation in the form

$$
\begin{aligned}
& 4 \cos x - \frac{\sqrt{3} \cos x}{\sin x} - 1 = 0 \Leftrightarrow 2 \sin x \cos x - \left(\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x\right) = 0, \\
& \sin 2 x - \left(\sin \frac{\pi}{3} \cos x + \cos \frac{\pi}{3} \sin x\right) = 0 \Leftrightarrow \sin 2 x - \sin \left(\frac{\pi}{3} + x\right) = 0 \Leftrightarrow \\
& \Leftrightarrow 2 \cos \left(\frac{3 x}{2} + \frac{\pi}{6}\right) \sin \left(\frac{x}{2} - \frac{\pi}{6}\right) = 0
\end{aligned}
$$

From this,

1) $2 \cos \left(\frac{3 x}{2} + \frac{\pi}{6}\right) = 0, \frac{3 x}{2} + \frac{\pi}{6} = \frac{\pi}{2} + \pi k, x_{1} = \frac{2 \pi}{9} + \frac{2}{3} \pi k = \frac{2}{9} \pi (3 k + 1), k \in \mathbb{Z}$;
2) $\sin \left(\frac{x}{2} - \frac{\pi}{6}\right) = 0, \frac{x}{2} - \frac{\pi}{6} = \pi n, x_{2} = \frac{\pi}{3} + 2 \pi n = \frac{\pi}{3} (6 n + 1), n \in \mathbb{Z}$.

Answer: $x_{1} = \frac{2}{9} \pi (3 k + 1), x_{2} = \frac{\pi}{3} (6 n + 1)$, where $k$ and $n \in \mathbb{Z}$.