3.367. $\frac{2 \sin 2 \alpha-3 \cos 2 \alpha}{4 \sin 2 \alpha+5 \cos 2 \alpha}$, if $\operatorname{tg} \alpha=3$.
Solution.

$\frac{2 \sin 2 \alpha-3 \cos 2 \alpha}{4 \sin 2 \alpha+5 \cos 2 \alpha}=\left[\sin x=\frac{2 \operatorname{tg} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}} ; \cos x=\frac{1-\operatorname{tg}^{2} \frac{x}{2}}{1+\operatorname{tg}^{2} \frac{x}{2}}\right]=$

$=\frac{\frac{4 \operatorname{tg} \alpha}{1+\operatorname{tg}^{2} \alpha}-\frac{3-3 \operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}}{\frac{8 \operatorname{tg} \alpha}{1+\operatorname{tg}^{2} \alpha}+\frac{5-5 \operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}}=\frac{4 \operatorname{tg} \alpha-3+3 \operatorname{tg}^{2} \alpha}{8 \operatorname{tg} \alpha+5-5 \operatorname{tg}^{2} \alpha}=$

$=[\operatorname{tg} \alpha=3]=\frac{4 \cdot 3-3+3 \cdot 9}{8 \cdot 3+5-5 \cdot 9}=-\frac{9}{4}$.

Answer: $-\frac{9}{4}$.

Knowing that $A, B, C$ are the interior angles of a certain triangle, prove the validity of the equalities (3.368-3.374):