Absolute value in trigonometric inequality $\left|\sin (2x)\right|\le \frac{\sqrt 2}2$ I know how to solve trigonometric equations and inequalities but I don't understand how to solve trigonometric inequalities with absolute value.
I find all the solution of the following inequality
$$\left|\sin (2x)\right|\le \frac{\sqrt 2}2$$
but I don't know what is the final solution.
I find $$( \frac{180}{8}, 3(\frac{180}{8}), 5(\frac{180}{8}), 7(\frac{180}{8}))$$ because I have to take $x$ that are between $[0,180]$.
But I don't know what to do from here.
 A: We have that
$$\sin (2x)=\frac{\sqrt 2}2 \implies 2x=\frac \pi 4+2k\pi \quad \lor \quad 2x= \frac34 \pi+2k\pi$$
$$\sin (2x)=-\frac{\sqrt 2}2 \implies 2x=-\frac \pi 4+2k\pi \quad \lor \quad 2x= \frac54 \pi+2k\pi$$
therefore we have
$$\left|\sin (2x)\right|\le \frac{\sqrt 2}2$$
for
$$-\frac \pi 4+2k\pi\le 2x \le \frac \pi 4+2k\pi \quad \lor \quad \frac34 \pi+2k\pi\le 2x \le \frac54 \pi+2k\pi$$
here is a sketch to visualize the solution for $2x$

and then in general we have
$$x \in \left[-\frac \pi 8+k\pi, \frac \pi 8+k\pi \right]\cup \left[\frac 3 8 \pi+k\pi, \frac 5 8 \pi+k\pi \right]$$
and for $x \in [0,\pi]$ we finally obtain
$$x \in \left[0, \frac \pi 8 \right]\cup \left[\frac 3 8 \pi, \frac 5 8 \pi \right]\cup \left[\frac 7 8 \pi,\pi \right]$$
A: Hint:
You can use that comparing the absolute values is the same as comparing the  squares:
$$|\sin 2x|\le\frac{\sqrt 2}2\iff\sin^22x=\frac{1-\cos 4x}2\le\frac12\iff\cos 4x\ge 0.$$
Now $\:\cos 4x\ge0\iff -\frac\pi 2\le 4x\le \frac\pi 2\mod 2\pi$ – meaning that
$$-\frac\pi 2+2k\pi\le 4x\le \frac\pi 2+2k\pi\quad \text{ for some }k\in\mathbf Z.$$
When you've finished the calculations, eliminate the solutions outside the  required interval.
A: If $x\in[0;180]$
$|\sin2x|\leq \dfrac{\sqrt{2}}{2}$
$ -\dfrac{\sqrt{2}}{2} \leq\sin2x\leq \dfrac{\sqrt{2}}{2}$
$0\leq2x\leq \dfrac{{\pi}}{4}$,
$\dfrac{3\pi}{4}\leq2x\leq {\dfrac{5\pi}{4}}$,
$\dfrac{7\pi}{4}\leq2x\leq {2\pi}$,
$\left[0;\dfrac{{\pi}}{8}\right] \cup \left[\dfrac{{3\pi}}{8};\dfrac{{5\pi}}{8}\right]\cup \left[\dfrac{{7\pi}}{8};{\pi}\right]$
A: I find this solution but I don't understand you find something else
$$S=\left(0, \frac \pi 8 \right]\cup \left[\frac 3 8 \pi, \frac 5 8 \pi \right]\cup \left[\frac 7 8 \pi,\pi \right)$$
