7. Given $\operatorname{tg} 2 \alpha=\frac{3}{4}, \alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, when the function $f(x)=\sin (x+2)+$ $\sin (\alpha-x)-2 \sin \alpha$ has a minimum value of 0, find the values of $\cos 2 \alpha$ and $\operatorname{tg} \frac{\alpha}{2}$.
7. From $\operatorname{tg} 2 \alpha=\frac{3}{4}, \alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we get $\operatorname{tg} \alpha=-3$ or $\operatorname{tg} \alpha=\frac{1}{3}$.

Given the function $f(x)=2 \sin \alpha(\cos x-1)$, when $\cos x=1$, $f(x)=0$ is the minimum value. Therefore, $\sin \alpha<0$, which means $-\frac{\pi}{2}<\alpha<0$, hence $\operatorname{tg} \alpha=-3$.
Thus, $\cos 2 \alpha=-\frac{4}{5}, \sin \alpha=-\frac{3 \sqrt{10}}{10}, \operatorname{tg} \frac{\alpha}{2}=\frac{1-\sqrt{10}}{3}$.