4.87 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) $\frac{7 \pi}{4}<\alpha<2 \pi$
4.87 Solving the quadratic equation with respect to $\operatorname{ctg} \alpha$, we find $(\operatorname{ctg} \alpha)_{1}=-\frac{1}{2}\left((\operatorname{tg} \alpha)_{1}=-2\right)$ and $(\operatorname{ctg} \alpha)_{2}=-3\left((\operatorname{tg} \alpha)_{2}=-\frac{1}{3}\right)$.

Now, using the condition of the problem, we get the answer:

a) $\left.(\cos 2 \alpha)_{1}=-\frac{3}{5} ; \sigma\right)(\cos 2 \alpha)_{2}=\frac{4}{5}$.