note that $\sum_{k=0}^n C_n^k\cos2kx= \mbox{Re}(\sum_{k=0}^n C_n^ke^{2kxi}) = \mbox{Re}[(e^{2xi}+1)^n]$. now we wish to write $e^{2xi}+1 = 1+\cos 2x+i\sin 2x$ in trigonometric form... its norm is $\sqrt{2+2\cos 2x}$, and its argument, $\gamma$, is such that $\cos\gamma = \frac{1+\cos 2x}{\sqrt{2+2\cos 2x}} = \cos\frac{ 2x}{2} = \pm \cos x$. so if $x$ ends in $[-\pi/2, \pi/2]$, we take $\gamma = x$ and if $x$ ends in $(\pi/2, 3\pi/2)$ we take $\gamma = x+\pi$. (because in all cases $\sqrt{2+2\cos 2x}\cos\gamma = 1+\cos 2x$ must be $\geq 0$.) now we have $\mbox{Re}[(e^{2xi}+1)^n] = \sqrt{2+2\cos\2x}^n\cos n\gamma$. if $n$ is even, then the original equation becomes $\sqrt{2+2\cos\2x}^n\cos nx = \cos nx$. then either $\cos nx = 0$, in which case $x = \frac{(2k+1)\pi}{2n}, \ k\in \mathbb Z$, or $2+2\cos\2x = 1$, in which case $\cos 2x = -\frac12$, i.e. $x = k\pi \pm \frac{\pi}{3}, \ k\in \mathbb Z$. if $n$ is odd and $x$ ends in $(\pi/2, 3\pi/2)$, the original equation becomes $\sqrt{2+2\cos\2x}^n\cos nx = -\cos nx$. if $\cos nx \neq 0$ then we get $\sqrt{2+2\cos\2x}^n = -1$ which is impossible. otherwise, $\cos nx = 0$, in which case $x = \frac{(2k+1)\pi}{2n}, \ k \in \mathbb Z$. (all these are solutions, whether they end in the presumed interval or not...) if $n$ is odd and $x$ ends in $[-\pi/2, \pi/2]$, the original equation becomes $\sqrt{2+2\cos\2x}^n\cos nx = \cos nx$, in which case either $\cos nx = 0$, in which case we have the solutions mentioned above or $\cos 2x = -\frac12$, i.e. $x = 2k\pi \pm \frac{\pi}{3}$ (because we assumed $x$ ends in $[-\pi/2, \pi/2]$)

Nice solution! pleurestique