Question

The number of values of a in [$0,2\pi$] for which 2 $si{n}^3$ $\alpha -7$ $si{n}^2$ $\alpha +7$ sin $\alpha =2$, is :

• A. 6
• B. 4
• C. 3
• D. 1

Answer 1

Answer: (C)

$2{\sin }^3\alpha -2=7{\sin }^2\alpha -7\sin \alpha$
$2\left(\sin \alpha -1\right)\left({\sin }^2\alpha +1+\sin \alpha \right)=7\sin \alpha \left(\sin \alpha -1\right)$
$\Rightarrow \sin \alpha =1$ or
$2si{n}^2\alpha +2+2\sin \alpha =7\sin \alpha \left(\sin \alpha -t\right)\left(\sin \alpha -2\right)$
$\Rightarrow \sin \alpha =1$ or $\sin \alpha =\frac{1}{2}\because \sin \alpha \ne -2$
$\Rightarrow 3$ solutions