Question

No of solutions of $1+\sin x{\sin }^2\frac{x}{2}=0$

Answer 1

$1+\sin x{\sin }^2\frac{x}{2}=0$
or $2+2\sin x{\sin }^2\frac{x}{2}=0$
or $2+\sin x(1-\cos x)=0$
or $4+2\sin x-\sin 2x=0$
or $\sin 2x=2\sin x+4$
The above result 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.
Ans: 0