Question

The number of solutions of the equation $1+\sin x{\sin }^2\frac{x}{2}=0,$ in $[-\pi ,\pi ]$

• A. zero
• B. one
• C. two
• D. three

Answer 1

The correct option is A zero

Given that, $1+\sin x{\sin }^2\frac{x}{2}=0$
$\therefore 1+\sin x\left(\frac{1-\cos x}{2}\right)=0$
$\Rightarrow 2+\sin x–\sin x\cos x=0$
$\Rightarrow \sin 2x–2\sin x=4$
This is not possible for any x given the fact that range of sine function is [-1,1].