The number of solutions of the equation $1 + \sin x\sin^2 {\frac{x}{2}} =0$ in $[-\pi,\pi]$ is...
What I have tried...
Since, $\sin \frac{x}{2} = \sqrt{\frac{1-\cos x}{2}}$,
$$ 1 + \sin x\cdot\frac{1-\cos x}{2} = 0$$
$$2+\sin x \cdot (1-\cos x) = 0$$
From here onwards I am not sure how to continue...
P.S. The answer to this question is $0$ ( which means that the equation has no solution. Please explain how) Thank you!
Hint. Don't try to solve the equation algebraicly. Instead, the product of your sine terms has to be $-1$. What are the possible outputs of sin? How can you get that product to be $-1$?