Question

The number of solution of the equation $1+\sin x{\sin }^2\left(\frac{x}{2}\right)=0,-\pi ,\pi$ is -

• A. Zero
• B. One
• C. Two
• D. Three

Answer 1

The correct option is B One

Given that, $1+\sin x{\sin }^2\left(\frac{x}{2}\right)=0$
Therefore, $1+\sin x\left[\frac{1-\cos x}{2}\right]=0$
$\Rightarrow 2+\sin x-\sin x\cos x=0$

$\Rightarrow 2+\sin x-\frac{\sin 2x}{2}=0$
$\Rightarrow \sin 2x-2\sin x=4$
Since, the maximum values of $\sin x$ and $\sin 2x$ are $1$, which is not possible for any $x$ in $[-\pi ,\pi ]$.