Question

If $f\left(x\right)=\frac{\sin \left(2nx\right)}{1+{\cos }^2\left(nx\right)},n\in N$ has $\frac{\pi }{6}$ as its fundamental period, then $n$ is equal to

Answer 1

$f\left(x\right)=\frac{\sin \left(2nx\right)}{1+{\cos }^2\left(nx\right)}=\frac{\sin \left(2nx\right)}{1+\frac{1+\cos \left(2nx\right)}{2}}$
$=\frac{2\sin \left(2nx\right)}{3+\cos \left(2nx\right)}$
Fundamental period of $\sin \left(2nx\right)$ is $\frac{2\pi }{2n}=\frac{\pi }{n}$
Fundamental period of $3+\cos \left(2nx\right)$ is $\frac{2\pi }{2n}=\frac{\pi }{n}$
Hence, period of $f\left(x\right)$ is $L.C.M.(\frac{\pi }{n},\frac{\pi }{n})=\frac{\pi }{n}$

Let us check $f(x+\frac{\pi }{2n})=\frac{2\sin \left[2n(\frac{\pi }{2n}+x)\right]}{3+\cos \left[2n(\frac{\pi }{2n}+x)\right]}$
$=\frac{-2\sin \left(2nx\right)}{3-\cos \left(2nx\right)}\ne f\left(x\right)$
Hence, fundamental period of $f\left(x\right)$ is $\frac{\pi }{n}.$
$\Rightarrow n=6$