Answer
$x=\frac{\pi}{6}+n\pi$ and $x=\frac{5\pi}{6}+n\pi$, where $n$ is an integer.
Work Step by Step
$4~cos^2x-1=0$
$4~cos^2x=1$
$cos^2x=\frac{1}{4}$
$cos~x=±\frac{1}{2}$
The period of $cos~x$ is $2\pi$. The solutions in the interval: $[0,2\pi)$ are:
$x=\frac{\pi}{6}$, $x=\frac{5\pi}{6}$, $x=\frac{7\pi}{6}$ and $x=\frac{11\pi}{6}$
Now, add multiples of $2\pi$ to each of the solutions:
$x=\frac{\pi}{6}+2n\pi$, $x=\frac{5\pi}{6}+2n\pi$, $x=\frac{7\pi}{6}+2n\pi$ and $x=\frac{11\pi}{6}+2n\pi$
But, notice that $x=\frac{7\pi}{6}=\frac{\pi}{6}+\pi$ and $x=\frac{11\pi}{6}=\frac{5\pi}{6}+\pi$
We can rewrite the solutions as:
$x=\frac{\pi}{6}+n\pi$ and $x=\frac{5\pi}{6}+n\pi$, where $n$ is an integer.
