Find all roots of the equation :$(1+\frac{ix}n)^n = (1-\frac{ix}n)^n$

Hint: Put $$z=\frac{1+i\frac{x}{n}}{1-i\frac{x}{n}}$$ then $z$ will be a $n$-root of unity and solve for $x:$ $$z= \frac{1+i\frac{x}{n}}{1-i\frac{x}{n}}=\exp{\left(i\frac{2k\pi}{n}\right)},\quad k\in\{0,1,...,n-1\}$$

Let $1=r\cos t,\dfrac xn=r\sin t$

Using http://mathworld.wolfram.com/deMoivresIdentity.html

$ r^n(\cos nt+i\sin nt)=r^n(\cos nt-i\sin nt)$

$\iff \sin nt=0$

$\implies nt=k\pi$

$ \dfrac xn=\tan t=\tan\dfrac{k\pi}n$ where $0\le k\le n-1$