Question

The general solution of the equation, $\frac{1-\sin x+...+{(-1)}^n{\sin }^{n}x+...}{1+\sin x+...+{\sin }^{n}x+...}=\frac{1-\cos 2x}{1+\cos 2x}$ is:

• A. ${(-1)}^n\frac{\pi }{3}+n\pi$
• B. ${(-1)}^n\frac{\pi }{6}+n\pi$
• C. ${(-1)}^{n+1}\frac{\pi }{6}+n\pi$
• D. ${(-1)}^{n-1}\frac{\pi }{3}+n\pi$

Answer 1

The correct option is B ${(-1)}^n\frac{\pi }{6}+n\pi$

$\frac{1-\sin x+...+{(-1)}^n{\sin }^{n}x+...}{1+\sin x+...+{\sin }^{n}x+...}$=$\frac{1-\cos 2x}{1+\cos 2x}$
$\frac{\frac{1}{1+\sin x}}{\frac{1}{1-\sin x}}=\frac{2{\sin }^{2}x}{2{\cos }^{2}x}$
$\frac{1-\sin x}{1+\sin x}=\frac{{\sin }^{2}x}{{\cos }^{2}x}$
$\Rightarrow 2{\sin }^{2}x+\sin x-1=0$
$\Rightarrow (2\sin x-1)(\sin x+1)=0$
$\Rightarrow \sin x=-1or\sin x=\frac{1}{2}$
$\Rightarrow \sin x=\sin \frac{3\pi }{2}or\sin x=\sin \frac{\pi }{6}$
The general solutions are
$x=m\pi +{(-1)}^m\frac{3\pi }{2},m\in z,x=n\pi +{(-1)}^n\frac{\pi }{6}$