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Q.
The number of solutions of the equation $1+\sin x \cdot \sin ^{2} \frac{x}{2}=0$ in $[-\pi, \pi]$ is
ManipalManipal 2013
Solution:
$1+\sin x \cdot \sin ^{2} \frac{x}{2}=0$
$\Rightarrow 2+2 \sin x \cdot \sin ^{2} \frac{x}{2}=0$
$\Rightarrow 2+\sin x(1-\cos x)=0$
$\Rightarrow 4+2 \sin x(1-\cos x)=0$
$\Rightarrow 4+2 \sin x-\sin 2 x=0$
$\Rightarrow \sin 2 x=2 \sin x+4$
Above is not possible for any value of $x$ as LHS has maximum value $1$ and RHS has value $2$.
Hence, there is no solution.