Q. The number of solutions of the equation $1 + \sin{x\sin^{2}{\frac{x}{2} = 0}}$ in $\lbrack - \pi,\pi\rbrack$ is
A) zero
B) one
C) two
D) three
Explanation
(a)
Since, $1 + \sin{x\sin^{2}{\frac{x}{2} = 0}}$
$\therefore\ 1 + \sin{x\left( \ \frac{1 - \cos x}{2} \right) = 0}$
$\Rightarrow \ 2 + \sin{x - \sin{x\cos x}} = 0$
$\Rightarrow \ \sin{2x - 2\sin{x = 4}}$
Which is not possible for any $x$ in $\lbrack - \pi,\ \pi\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.