Question

The general solution of the equation $1+\sin x{\sin }^2\frac{x}{2}=0$ is (where $n\in Z$)

• A. $n\pi +(-1{)}^n\frac{\pi }{2}$
• B. $n\pi +(-1{)}^n\frac{\pi }{4}$
• C. $n\pi +(-1{)}^n\frac{\pi }{6}$
• D. No solution

Answer 1

The correct option is D No solution

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