2. Given: $\frac{1-4 \sin \alpha}{\tan \alpha}=\sqrt{3}, \alpha \in\left(0, \frac{\pi}{2}\right)$, then $\alpha=$ $\qquad$
Obviously, $\sin \alpha<\frac{1}{4} \Rightarrow \alpha \in\left(0, \frac{\pi}{6}\right)$, then $\frac{1-4 \sin \alpha}{\tan \alpha}=\sqrt{3} \Rightarrow \cos \alpha-4 \sin \alpha \cos \alpha=\sqrt{3} \sin \alpha$ $\Rightarrow \sin 2 \alpha=\frac{1}{2} \cos \alpha-\frac{\sqrt{3}}{2} \sin \alpha=\sin \left(\frac{\pi}{6}-\alpha\right) \Rightarrow 2 \alpha=\frac{\pi}{6}-\alpha \Rightarrow \alpha=\frac{\pi}{18}$.