$2 \sin ^{3} \alpha-7 \sin ^{2} \alpha+7 \sin \alpha-2=0$

$\Rightarrow 2 \sin ^{2} \alpha(\sin \alpha-1)-5 \sin \alpha$

$(\sin \alpha-1)+2(\sin \alpha-1)=0$

$\Rightarrow(\sin \alpha-1)\left(2 \sin ^{2} \alpha-5 \sin \alpha+2\right)$ $=0$

$\Rightarrow \sin \alpha-1=0$ or $2 \sin ^{2} \alpha-5 \sin \alpha+$ $2=0$

$\sin \alpha=1$ or $\sin \alpha=\frac {5 \pm \sqrt{25-16}} {4}=\frac{5 \pm 3}{4}$

$\alpha=\frac{\pi}{2}$

or $\sin \alpha=\frac{1}{2}, 2$

Now, $\sin \alpha \neq 2$

for, $\sin \alpha=\frac{1}{2}$

$\alpha=\frac{\pi}{3}, \frac{2 \pi}{3}$

There are three values of $\alpha$ between $[0,2 \pi]$