Example 1 (2002 National College Entrance Examination) Given $\sin ^{2} 2 \alpha+\sin 2 \alpha \cos \alpha-\cos 2 \alpha=1, \alpha \in\left(0, \frac{\pi}{2}\right)$, find the values of $\sin \alpha$ and $\tan \alpha$.
Analyze: Use trigonometric formulas to eliminate 1, and transform the relation into a relation of angle $\alpha$, and solve further.

Solution: By the double-angle formulas $\sin 2 \alpha=2 \sin \alpha \cos \alpha, \cos 2 \alpha=2 \cos ^{2} \alpha-1$, the original equation can be transformed into: $4 \sin ^{2} \alpha \cos ^{2} \alpha+2 \sin \alpha \cos ^{2} \alpha-2 \cos ^{2} \alpha=0$
That is
$$
\begin{array}{l}
\cos ^{2} \alpha\left(2 \sin ^{2} \alpha+\sin \alpha-1\right)=0 \\
\cos ^{2} \alpha(2 \sin \alpha-1)(\sin \alpha+1)=0
\end{array}
$$

Since $\alpha \in\left(0, \frac{\pi}{2}\right)$, we have $\sin \alpha+1 \neq 0, \cos \alpha \neq 0$
Therefore, $2 \sin \alpha-1=0$, so $\sin \alpha=\frac{1}{2}, \alpha=\frac{\pi}{6}$, so $\tan \alpha=\frac{\sqrt{3}}{3}$. In summary: $\sin \alpha=\frac{1}{2}, \tan \alpha=\frac{\sqrt{3}}{3}$