Question

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

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

Answer 1

Answer: (B)

$\frac{1-\sin x+\dots +(-1{)}^n{\sin }^{n}x+\dots \dots }{1+\sin x+\dots +{\sin }^{n}x+\dots }=\frac{1-\cos 2x}{1+\cos 2x}$
$\Rightarrow \frac{1}{1+\sin x}\times \frac{1-\sin x}{1}=\frac{2{\sin }^{2}x}{2{\cos }^{2}x}$
$\Rightarrow {\cos }^{2}x-{\cos }^{2}x\sin x={\sin }^{2}x+{\sin }^{3}x(\sin x+1\ne 0)$
$\Rightarrow 2{\sin }^{2}x+\sin x-1=0$
$\Rightarrow \sin x=-1$ or $\sin x=1/2$
Since $\sin x\ne -1$
we have $\sin x=1/2=\sin (\pi /6)$
$\Rightarrow x=n\pi +(-1{)}^n(\pi /6),n\in Z$