8.039. $3 \sin 2 x+2 \cos 2 x=3$.

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

8.039. $3 \sin 2 x+2 \cos 2 x=3$.
Solution.

We have

$6 \sin x \cos x+2\left(\cos ^{2} x-\sin ^{2} x\right)=3\left(\cos ^{2} x+\sin ^{2} x\right)$,

$5 \sin ^{2} x-6 \sin x \cos x+\cos ^{2} x=0, \Leftrightarrow 5 \operatorname{tg}^{2} x-6 \operatorname{tg} x+1=0$.

Solving this equation as a quadratic equation in terms of $\operatorname{tg} x$, we find $(\operatorname{tg} x)_{1}=\frac{1}{5}$ or $(\operatorname{tg} x)_{2}=1$, from which $x_{1}=\operatorname{arctg} \frac{1}{5}+\pi k=\operatorname{arcctg} 5+\pi k, k \in Z$; $x_{2}=\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n+1), n \in Z$.

Answer: $\quad x_{1}=\operatorname{arcctg} 5+\pi k ; x_{2}=\frac{\pi}{4}(4 n+1), \quad k, n \in Z$.