2. Given $\frac{\sin ^{3} \alpha}{\sin \left(\frac{\pi}{6}-3 \alpha\right)}=\frac{\cos ^{3} \alpha}{\cos \left(\frac{\pi}{6}-3 \alpha\right)}$ $=\sqrt{3}\left(\alpha \neq \frac{k \pi}{4}, k \in Z\right)$. Then the value of $\operatorname{tg} 2 \alpha$ is
2. $\frac{\sqrt{3}}{4}$.

From the given, we have
(1) + (2), we get
$$
\begin{array}{l}
\frac{\sqrt{3} \sin \left(\frac{\pi}{6}-3 \alpha\right)}{\sin \alpha}+\frac{\sqrt{3} \cos \left(\frac{\pi}{6}-3 \alpha\right)}{\cos \alpha}=1 . \\
\therefore \sqrt{3} \sin \left(\frac{\pi}{6}-3 \alpha\right) \cos \alpha+\sqrt{3} \cos \left(\frac{\pi}{6}-3 \alpha\right) \sin \alpha \\
=\sin \alpha \cos \alpha,
\end{array}
$$

which means $\sqrt{3} \sin \left(\frac{\pi}{6}-2 \alpha\right)=\frac{1}{2} \sin 2 \alpha$,
or $\sqrt{3} \cos 2 \alpha=4 \sin 2 \alpha$.
$$
\therefore \operatorname{tg} 2 \alpha=\frac{\sqrt{3}}{4} \text {. }
$$