3. Let $0^{\circ}<\alpha<90^{\circ}$. If $1+\sqrt{3} \operatorname{tg}(60-\alpha)$ $=\frac{1}{\sin \alpha}$, then $\alpha$ equals $\qquad$.
3. $30^{\circ}$ or $50^{\circ}$.

The original equation is equivalent to the following three equations:
$$
\begin{array}{l}
\sin \alpha\left(1+\sqrt{3} \cdot \frac{\sqrt{3}-\tan \alpha}{1+\sqrt{3} \tan \alpha}\right)=1, \\
\sin \alpha(1+\sqrt{3} \tan \alpha+3-\sqrt{3} \tan \alpha) \\
=1+\sqrt{3} \tan \alpha, \\
4 \sin \alpha=\sqrt{3} \tan \alpha+1, \\
2 \sin 2 \alpha=\sqrt{3} \sin \alpha+\cos \alpha=2 \sin \left(\alpha+30^{\circ}\right),
\end{array}
$$

Thus, $\sin 2 \alpha=\sin \left(\alpha+30^{\circ}\right)$.
Therefore, either $2 \alpha=\alpha+30^{\circ}$, or $2 \alpha+\left(\alpha+30^{\circ}\right)=180^{\circ}$. Hence, $\alpha=30^{\circ}$ or $50^{\circ}$.