3.382. Find $\sin 2 \alpha$, if it is known that $2 \operatorname{tg}^{2} \alpha-7 \operatorname{tg} \alpha+3=0$ and the number $\alpha$ satisfies the inequalities: a) $\pi<\alpha<\frac{5 \pi}{4}$; b) $\frac{5 \pi}{4}<\alpha<\frac{3 \pi}{2}$.
## Solution.

Solving the equation $2 \operatorname{tg}^{2} \alpha-7 \operatorname{tg} \alpha+3=0$ as a quadratic equation in $\operatorname{tg} \alpha$, we get $(\operatorname{tg} \alpha)_{1,2}=\frac{7 \pm \sqrt{49-24}}{4}=\frac{7 \pm 5}{4}, \operatorname{tg} \alpha=\frac{1}{2}$ or $\operatorname{tg} \alpha=3$.

In case a) $\pi<\alpha<\frac{5 \pi}{4}, \operatorname{tg} \pi<\operatorname{tg} \alpha<\operatorname{tg} \frac{5 \pi}{4}, 0<\operatorname{tg} \alpha<1 \Rightarrow \operatorname{tg} \alpha=\frac{1}{2}$. Then $\sin 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1+\operatorname{tg}^{2} \alpha}=\frac{2 \cdot \frac{1}{2}}{1+\frac{1}{4}}=\frac{4}{5}$.

In case b) $\frac{5 \pi}{4}<\alpha<\frac{3 \pi}{2}, \operatorname{tg} \frac{5 \pi}{4}<\operatorname{tg} \alpha<\operatorname{tg} \frac{3 \pi}{2}, 1<\operatorname{tg} \alpha<\infty \Rightarrow$ $\Rightarrow \operatorname{tg} \alpha=3$. Then

$$
\sin 2 \alpha=\frac{2 \operatorname{tg} \alpha}{1+\operatorname{tg}^{2} \alpha}=\frac{2 \cdot 3}{1+9}=\frac{3}{5}
$$

Answer: a) $\frac{4}{5}$; b) $\frac{3}{5}$.