8.267. $\frac{2\left(\cos ^{3} x+2 \sin ^{3} x\right)}{2 \sin x+3 \cos x}=\sin 2 x$.
## Solution.

Domain of definition: $2 \sin x + 3 \cos x \neq 0$.

## Write the equation as

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

Let $\operatorname{tg} x = y$. We have

$$
\begin{aligned}
& 2 y^{3} - 2 y^{2} - 3 y + 1 = 0 \Leftrightarrow \\
& \Leftrightarrow \left(y^{3} + 1\right) + y^{2}(y + 1) - 3 y(y + 1) = 0 \Leftrightarrow \\
& \Leftrightarrow (y + 1)\left(y^{2} - y + 1\right) + y^{2}(y + 1) - 3 y(y + 1) = 0 \Leftrightarrow \\
& \Leftrightarrow (y + 1)\left(\dot{y}^{2} - y + 1 + y^{2} - 3 y\right) = 0, (y + 1)\left(2 y^{2} - 4 y + 1\right) = 0
\end{aligned}
$$

From this, $y + 1 = 0, y_{1} = -1 ; 2 y^{2} - 4 y + 1 = 0, y_{2} = 1 - \frac{\sqrt{2}}{2} ; y_{3} = 1 + \frac{\sqrt{2}}{2}$.

## We obtained

$\operatorname{tg} x = -1, x_{1} = -\frac{\pi}{4} + \pi k = \frac{\pi}{4}(4 k - 1), k \in Z$;

$\operatorname{tg} x = 1 - \frac{\sqrt{2}}{2}, x_{2} = \operatorname{arctg}\left(1 - \frac{\sqrt{2}}{2}\right) + \pi n, n \in Z ;$

$\operatorname{tg} x = 1 + \frac{\sqrt{2}}{2}, x_{3} = \operatorname{arctg}\left(1 + \frac{\sqrt{2}}{2}\right) + \pi l, l \in Z$.

Answer: $x_{1} = \frac{\pi}{4}(4 k - 1) ; x_{2} = \operatorname{arctg}\left(1 - \frac{\sqrt{2}}{2}\right) + \pi n ; x_{3} = \operatorname{arctg}\left(1 + \frac{\sqrt{2}}{2}\right) + \pi l$, where $k, n$ and $l \in Z$.