Question

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

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

Answer 1

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

Given, $\frac{1-\sin x+...+{(-1)}^n{\sin }^{n}x+...}{1+\sin x+...+{\sin }^{n}x}=\frac{1-\cos 2x}{1+\cos 2x}$

$\Rightarrow \frac{1}{1+\sin x}\times \frac{1-\sin x}{1}=\frac{2{\sin }^{2}x}{2{\cos }^{2}x}$ as $-1<\sin x<1$

$\Rightarrow 1-\sin x=\frac{{\sin }^{2}x(1+\sin x)}{1-{\sin }^{2}x}\Rightarrow (1-{\sin }^{2}x)={\sin }^{2}x$

$\Rightarrow 1-2\sin x=0\Rightarrow \sin x=\frac{1}{2}\Rightarrow n\pi +{(-1)}^n\left(\frac{\pi }{6}\right)$