Question

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

• A. $x=2n\pi +\frac{\pi }{3}$, $n\in Z$
• B. $x=2n\pi +\frac{\pi }{4}$, $n\in Z$
• C. $x=2n\pi +\frac{\pi }{2}$, $n\in Z$
• D. $x=2n\pi +\frac{\pi }{8}$, $n\in Z$

Answer 1

The correct option is C $x=2n\pi +\frac{\pi }{2}$, $n\in Z$

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

$\frac{(\sin \frac{x}{2}-\cos \frac{x}{2})(1+\sin \frac{x}{2}\cos \frac{x}{2})}{2(1+\sin \frac{x}{2}\cos \frac{x}{2})}=\frac{\cos x}{3}$

$\sin \frac{x}{2}-\cos \frac{x}{2}=\frac{2}{3}\cos x$

$1-\sin x=\frac{4}{9}{\cos }^{2}x$ (squaring)

$\frac{4}{9}{\sin }^{2}x-\sin x+1-\frac{4}{9}=0$

$\frac{4}{9}{\sin }^{2}x-\sin x+1-\frac{4}{9}=0$

$4{\sin }^{2}x-9\sin x+5=0$

$(4\sin x-5)(\sin x-1)=0$

$\sin x=1,(\sin x\ne \frac{5}{4})$ $\Rightarrow$

$x=2n\pi +\frac{\pi }{2}$, $nϵZ$