Yes it's true for the exact reason you posted. You could also try to solve the equation...

$\displaystyle \begin{align*} 2\cos^2{(4x)} - 1 &= 0 \\ 2\cos^2{(4x)} &= 1 \\ \cos^2{(4x)} &= \frac{1}{2} \\ \cos{(4x)} &= \pm \frac{1}{\sqrt{2}} \\ 4x &= \left\{ \frac{\pi}{4} , \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\} + 2\pi\,n \textrm{ where } n \in \mathbf{Z} \\ x &= \left\{ \frac{\pi}{16}, \frac{3\pi}{16}, \frac{5\pi}{16}, \frac{7\pi}{16} \right\} + \frac{\pi}{2}n \end{align*}$

Clearly $\displaystyle \begin{align*} x = \frac{\pi}{16} \end{align*}$ is a solution.