Question

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

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

Answer 1

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

the numerator and denominator of LHS are both GP's
so we find there sums using sum of GP formula
we get
$\frac{(1-(-\sin x{)}^{n+1}\left)\right(1-\sin x)}{(1-(\sin x{)}^{n+1}\left)\right(1+\sin x)}=\frac{1-\cos 2x}{1+\cos 2x}$
taking n as even , we conclude that there is no solution
take n as odd, we get
$\frac{1-\sin x}{1+\sin x}=\frac{1-\cos 2x}{1+\cos 2x}$
so $\sin x=\cos 2x=1-2{\sin }^{2}x$
this gives
$\sin x=-1,\frac{1}{2}$
we reject $-1$ because then $1+\sin x$ is not defined
so $\sin x=\frac{1}{2}$
so $x=n\pi +(-1{)}^n\frac{\pi }{6}$