The sum of all positive real solutions to the equation $2\cos 2x(\cos 2x-\cos \frac{36{\pi }^2}{x})=\cos 4x-1$ is $k\pi$, then the value of k is ___. 

$2{\cos }^{2}2x-2\cos 2x\cos \frac{36{\pi }^2}{x}=-2{\sin }^{2}2x$

$\Rightarrow \cos (2x+\frac{36{\pi }^2}{x})+\cos (2x-\frac{36{\pi }^2}{x})=2$

$\Rightarrow 2x+\frac{36{\pi }^2}{x}=2m\pi and2x-\frac{36{\pi }^2}{x}=2n\pi$

$\Rightarrow 2x=(m+n)\pi =p\pi ,\text{and}\frac{36{\pi }^2}{x}=(m-n)\pi =q\pi ,$ where

$p=m+n\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}q=m-n$

$\therefore 2x.\frac{36{\pi }^2}{x}=pq{\pi }^2$

$\therefore x\in \{\pi ,2\pi ,3\pi ,6\pi ,9\pi ,18\pi \}.$