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

The correct option is A Zero
Given, $1+\sin x{\sin }^2\frac{x}{2}=0$
As we know, $\cos x=1-2{\sin }^2\frac{x}{2}$

So, ${\sin }^2\frac{x}{2}=\frac{1-\cos x}{2}$
$\therefore 1+\sin x\left(\frac{1-\cos x}{2}\right)=0$
$\Rightarrow 2+\sin x–\sin x\cos x=0$

Multiply by $2$ both sides
$\Rightarrow 4+2\sin x–2\sin x\cos x=0$
$\Rightarrow \sin 2x–2\sin x=4$

Since, the maximum value which $\sin 2x\text{and}2\sin x$ can take is $1$.

Hence, this is not possible for any $x$ in $[-\pi ,\pi ]$.

Hence, the correct answer is Option a.