8.299. $\frac{\tan x + \cot x}{\cot x - \tan x} = 6 \cos 2x + 4 \sin 2x$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\operatorname{ctg} x-\operatorname{tg} x \neq 0, \\ \cos x \neq 0, \\ \sin x \neq 0 .\end{array}\right.$

## Write the equation in the form

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0564.jpg?height=177&width=604&top_left_y=515&top_left_x=154)

$$
\begin{aligned}
& \Leftrightarrow \frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x} \cdot \frac{\cos x \sin x}{\sin ^{2} x-\cos ^{2} x}-6 \cos 2 x-4 \sin 2 x=0 \Leftrightarrow \\
& \Leftrightarrow \frac{1}{\cos 2 x}-6 \cos 2 x-4 \sin 2 x=0 \Leftrightarrow 1-6 \cos ^{2} 2 x-4 \sin 2 x \cos 2 x=0, \\
& \cos ^{2} 2 x+\sin ^{2} 2 x-6 \cos ^{2} 2 x-4 \sin 2 x \cos 2 x=0, \\
& \sin ^{2} 2 x-4 \sin 2 x \cos 2 x-5 \cos ^{2} 2 x=0, \operatorname{tg}^{2} 2 x-4 \operatorname{tg} 2 x-5=0
\end{aligned}
$$

Solving this equation as a quadratic in $\operatorname{tg} 2 x$, we get

$$
\begin{aligned}
& \operatorname{tg} 2 x=-1, 2 x_{1}=-\frac{\pi}{4}+\pi k, x_{1}=-\frac{\pi}{8}+\frac{\pi k}{2}, k \in Z; \operatorname{tg} 2 x=5 \\
& 2 x_{2}=\operatorname{arctg} 5+\pi n, x_{2}=\frac{1}{2} \operatorname{arctg} 5+\frac{\pi n}{2}, n \in Z
\end{aligned}
$$

Answer: $x_{1}=-\frac{\pi}{8}+\frac{\pi k}{2} ; x_{2}=\frac{1}{2} \operatorname{arctg} 5+\frac{\pi n}{2}$, where $k$ and $n \in Z$.