3.167. Find the number $\alpha \in\left(\frac{\pi}{2}, \pi\right)$, if it is known that $\operatorname{tg} 2 \alpha=-\frac{12}{5}$.
Solution.

$$
\frac{2 \tan \alpha}{1-\tan^{2} \alpha}=-\frac{12}{5}, \quad \frac{2 \tan \alpha}{1-\tan^{2} \alpha}+\frac{12}{5}=0 \Rightarrow
$$

$\Rightarrow 6 \tan^{2} \alpha-5 \tan \alpha-6=0$, from which $(\tan \alpha)_{1}=\frac{3}{2}$, which does not fit the solution of the problem, since by condition the angle belongs to the 2nd quadrant and its tangent is negative, and $(\tan \alpha)_{2}=-\frac{2}{3}$.

From this, $\alpha=-\arctan \frac{2}{3}+\pi n, n \in Z$. Since $\alpha \in\left(\frac{\pi}{2}, \pi\right)$, then $\alpha=-\arctan \frac{2}{3}+\pi$

Answer: $\pi-\arctan \frac{2}{3}$.