Question

Number of solutions of the equation ${\sin }^{3}x\cos x+{\sin }^{2}x{\cos }^{2}x+\sin x{\cos }^{3}x=1$ in the interval $[0,2\pi ]$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

We have ${\sin }^{3}x\cos x+{\sin }^{2}x{\cos }^{2}x+\sin x{\cos }^{3}x=1\Rightarrow \sin x\cos x\left({\sin }^{2}x+\sin x\cos x+{\cos }^{2}x\right)=1$
$\Rightarrow \frac{\sin 2x}{2}\left(1+\frac{1}{2}\sin 2x\right)=1\Rightarrow \sin 2x(2+\sin 2x)=4\Rightarrow {\sin }^{2}2x+2\sin 2x-4=0$
$\therefore \sin 2x=\frac{-2\pm \sqrt{4+16}}{2}=-1\pm \sqrt{5}\text{ (Not possible) }$
As, $-1\le \sin 2x\le 1\forall x\in R$
Hence, number of solutions equals 0 .