The number of values of α in 0 ,   2 π for which 2 sin 3 α - 7 sin 2 α + 7 sin&#945…

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

$\Rightarrow 2\left({\sin }^3 \mathrm{\alpha }-1\right)-7 \sin \mathrm{\alpha } \left(\sin \mathrm{\alpha }-1\right)=0$

$\Rightarrow \left(\sin \mathrm{\alpha }-1\right)\left[2\left({\sin }^2 \mathrm{\alpha }+\sin \mathrm{\alpha }+1\right)-7 \sin \mathrm{\alpha }\right]=0$ $\left\{\because a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right\}$

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

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

 $\therefore \sin \mathrm{\alpha }=1, \sin \mathrm{\alpha }=\frac{1}{2}, \sin \mathrm{\alpha }\ne 2, \because \mathrm{sinx}\in \left[-1,1\right]$ 
Now, $\sin \mathrm{\alpha }=1 \Rightarrow \mathrm{\alpha }=\frac{\mathrm{\pi }}{2} \left(\because \mathrm{\alpha }\in \left[\mathrm{0}, 2\mathrm{\pi }\right]\right)$
$\sin \mathrm{\alpha }=\frac{1}{2} \Rightarrow \mathrm{\alpha }=\frac{\mathrm{\pi }}{6}, \frac{5\mathrm{\pi }}{6} \left(\because \mathrm{\alpha } \in \left[\mathrm{0}, 2\mathrm{\pi }\right]\right)$

Hence, $3$ values of $\alpha$ in $\left[0\text{, }2\pi \right]$.