12. Let $x \in(0, \pi)$ and $\cot \frac{x}{3}=m \cot x, m$'s range of values is $\qquad$
12. $m>3$ or $m\cot \frac{\pi}{3}$, thus $t^{2}>\frac{1}{3}$, i.e., $\frac{3 m-1}{m-3}>\frac{1}{3}$. Therefore, $m>3$ or $m<0$.