Question

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

• A. zero
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A zero

Given, $1+\sin x\cdot {\sin }^2\frac{x}{2}=0$
$\Rightarrow 2+2\sin x\cdot {\sin }^2\frac{x}{2}=0$
$\Rightarrow 2+\sin x(1-\cos x)=0$
$\Rightarrow 4+2\sin x(1-\cos x)=0$
$\Rightarrow 4+2\sin x-\sin 2x=0$
$\Rightarrow \sin 2x=2\sin x+4$
Above is not possible for any value of $x$ as LHS has maximum value $1$ and RHS has minimum value $2$.
Hence, there is no solution.