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

Solving the equation $2 \operatorname{ctg}^{2} \alpha+7 \operatorname{ctg} \alpha+3=0$ as a quadratic equation in $\operatorname{ctg} \alpha$, we have $(\operatorname{ctg} \alpha)_{1,2}=\frac{-7 \pm \sqrt{49-24}}{4}=\frac{-7 \pm 5}{4}$, from which $\operatorname{ctg} \alpha=-3$, $\operatorname{tg} \alpha=-\frac{1}{3}$ or $\operatorname{ctg} \alpha=-\frac{1}{2}, \operatorname{tg} \alpha=-2$. In case a) $\frac{3 \pi}{2}<\alpha<\frac{7 \pi}{4}$, $\operatorname{tg} \frac{3 \pi}{2}<\operatorname{tg} \alpha<\operatorname{tg} \frac{7 \pi}{4},-\infty<\operatorname{tg} \alpha<-1 \Rightarrow \operatorname{tg} \alpha=-2$. Then

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

In case b) $\frac{7 \pi}{4}<\alpha<2 \pi, \operatorname{tg} \frac{7 \pi}{4}<\operatorname{tg} \alpha<\operatorname{tg} 2 \pi,-1<\operatorname{tg} \alpha<0 \Rightarrow$ $\Rightarrow \operatorname{tg} \alpha=-\frac{1}{3}$. Then
$\cos 2 \alpha=\frac{1-\operatorname{tg}^{2} \alpha}{1+\operatorname{tg}^{2} \alpha}=\frac{1-\frac{1}{9}}{1+\frac{1}{9}}=\frac{8}{10}=\frac{4}{5}$.

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