Question

Solve: $\frac{{\sin }^3\frac{x}{2}-{\cos }^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$

• A. $x=(2n+1)\frac{\pi }{2},nϵz$
• B. $x=(4n+1)\frac{\pi }{2},nϵz$
• C. $x=(4n+1)\frac{\pi }{4},nϵz$
• D. $x=(2n+1)\frac{\pi }{4},nϵz$

Answer 1

The correct option is B $x=(4n+1)\frac{\pi }{2},nϵz$

Given, $\frac{{\sin }^3\frac{x}{2}-{\cos }^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$
$\Rightarrow \frac{(\sin \frac{x}{2}-\cos \frac{x}{2})({\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+\sin \frac{x}{2}\cos \frac{x}{2})}{2+\sin x}=\frac{\cos x}{3}$
$\Rightarrow \frac{(\sin \frac{x}{2}-\cos \frac{x}{2})(1+\frac{\sin x}{2})}{2+\sin x}=\frac{\cos x}{3}$
$\Rightarrow 3(\sin \frac{x}{2}-\cos \frac{x}{2})=2({\sin }^2\frac{x}{2}-{\cos }^2\frac{x}{2})=2(\sin \frac{x}{2}-\cos \frac{x}{2})(\sin \frac{x}{2}+\cos \frac{x}{2})$
$\Rightarrow (\sin \frac{x}{2}-\cos \frac{x}{2})(\sin \frac{x}{2}+\cos \frac{x}{2}-\frac{3}{2})=0$
$\Rightarrow (\sin \frac{x}{2}-\cos \frac{x}{2})=0$ as $(\sin \frac{x}{2}+\cos \frac{x}{2}-\frac{3}{2})\ne 0$
$\Rightarrow \tan \frac{x}{2}=1\Rightarrow \frac{x}{2}=n\pi +\frac{\pi }{4},nϵz$
$\Rightarrow x=(4n+1)\frac{\pi }{2}$