Intersecting angles on trigonometric circle after solving

$|\sin(2 x - \pi/12)|\le\sqrt(2)/2$

I get two interval solutions which intersect eachother, 


$5 \pi/12 + k \pi \le x \le \pi/6 + k \pi$
and
$11\pi/12 + k \pi \le x\le 2\pi/3 + k \pi$


Since the solutions I get arent the basic, I don't have a picture in my mind how they look and intersect.
Note that $\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$. Also note that when $\sin$ is negative, the angle is also. So that means $$-\frac{\pi}{4}\le2x-\frac{\pi}{12}\le\frac{\pi}{4}$$ 

Can you solve it from here?
$$\left|\sin\left(2x-\dfrac\pi{12}\right)\right|\le\dfrac{\sqrt2}2$$

$$\iff2\sin^2\left(2x-\dfrac\pi{12}\right)\le1$$

Using $2\cos2A=1-2\sin^2A$

$$\iff\cos\left(4x-\dfrac\pi6\right)\ge0$$

$$\implies2m\pi-\dfrac\pi2\le4x-\dfrac\pi6\le2m\pi+\dfrac\pi2$$ where $m$ is any integer