Answer
$x=\frac{\pi}{4}+n·\frac{\pi}{2}$, where $n$ is an integer.
Work Step by Step
$2-4~sin^2x=0$
$2=4~sin^2x$
$sin^2x=\frac{1}{2}$
$sin~x=±\sqrt {\frac{1}{2}}=±\frac{\sqrt 2}{2}$
The period of $sin~x$ is $2\pi$. The solutions in the interval: $[0,2\pi)$ are:
$x=\frac{\pi}{4}$, $x=\frac{3\pi}{4}$, $x=\frac{5\pi}{4}$, $x=\frac{7\pi}{4}$
Notice that:
$x=\frac{3\pi}{4}=\frac{\pi}{4}+1·\frac{\pi}{2}$,
$x=\frac{5\pi}{4}=\frac{\pi}{4}+2·\frac{\pi}{2}$,
$x=\frac{7\pi}{4}=\frac{\pi}{4}+3·\frac{\pi}{2}$
Now, add multiples of $2\pi$ to each of the solutions:
$x=\frac{\pi}{4}+2n\pi$
$x=\frac{3\pi}{4}=\frac{\pi}{4}+\frac{\pi}{2}+2n\pi$,
$x=\frac{5\pi}{4}=\frac{\pi}{4}+2·\frac{\pi}{2}+2n\pi$,
$x=\frac{7\pi}{4}=\frac{\pi}{4}+3·\frac{\pi}{2}+2n\pi$
We can rewrite all the solutions as:
$x=\frac{\pi}{4}+n·\frac{\pi}{2}$, where $n$ is an integer.
