Question

The number of solutions of $\frac{{\sin }^3\frac{x}{2}-{\cos }^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$ in the interval $[0,10\pi ]$ is

• A. $2$
• B. $4$
• C. $6$
• D. $5$

Answer 1

The correct option is D $5$

Given, $\frac{{\sin }^3\frac{x}{2}-{\cos }^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$

$\Rightarrow \frac{(\sin \frac{x}{2}-\cos \frac{x}{2})(1+\sin \frac{x}{2}\cos \frac{x}{2})}{2(1+\sin \frac{x}{2}\cos \frac{x}{2})}=\frac{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}{3}\Rightarrow 3(\sin \frac{x}{2}-\cos \frac{x}{2})+2({\sin }^2\frac{x}{2}-{\cos }^2\frac{x}{2})=0\Rightarrow (\sin \frac{x}{2}-\cos \frac{x}{2})(3+2\sin \frac{x}{2}+2\cos \frac{x}{2})=0\Rightarrow \sin \frac{x}{2}=\cos \frac{x}{2}(\because \sin \theta +\cos \theta \in [-\sqrt{2},\sqrt{2}\left]\right)\Rightarrow \tan \frac{x}{2}=1\Rightarrow \frac{x}{2}=n\pi +\frac{\pi }{4},n\in Z\Rightarrow x=2n\pi +\frac{\pi }{2}$

In $[0,10\pi ],$
$x=\frac{\pi }{2},\frac{5\pi }{2},\frac{9\pi }{2},\frac{13\pi }{2},\frac{17\pi }{2}$
$\therefore$ Number of solutions is $5.$