Double angle formula equation $\sin(2x)\cos(2x) = \frac{1}{4}$

Question

For $0 \le x \lt 2\pi$. I have so far:

$$2\sin{x}\cos{x}(2\cos^2-1)-\frac{1}{4}= 0$$

I need help with the next step.

Answer 1

Multiplying by $2$ the equation becomes $\sin(4x)=\frac 1 2$ so $4x=\frac {\pi} 6$ and $x=\frac {\pi} {24}$ is one sultion. Can you write down all solutions with $0 \leq x <2\pi$? [There are eight of them!].