# 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.

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

$$x \in \left[0, \frac \pi 8 \right]\cup \left[\frac 3 8 \pi, \frac 5 8 \pi \right]\cup \left[\frac 7 8 \pi,\pi \right]$$

• Thanks but I have to white the solution as a section or as a union of sections. – Rony Cohen Oct 11 '20 at 13:57
• We can write it as $$x \in \left[-\frac \pi 8+k\pi, \frac \pi 8+k\pi \right]\cup \left[\frac 3 8 \pi+k\pi, \frac 5 8 \pi+k\pi \right]$$ – user Oct 11 '20 at 14:01

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.

I find this solution but I don't understand you find something else

$$S=\left(0, \frac \pi 8 \right]\cup \left[\frac 3 8 \pi, \frac 5 8 \pi \right]\cup \left[\frac 7 8 \pi,\pi \right)$$

• I found the same, but generalised. – Bernard Oct 11 '20 at 14:07
• It seems that $0$ and $\pi$ are included. – user Oct 11 '20 at 14:12
• It is almost right, it should be $$S=\left[0, \frac \pi 8 \right]\cup \left[\frac 3 8 \pi, \frac 5 8 \pi \right]\cup \left[\frac 7 8 \pi,\pi \right]$$ – user Oct 11 '20 at 14:16
• You should include your solution editing the original question. – user Oct 11 '20 at 14:16

If $$x\in[0;180]$$

$$|\sin2x|\leq \dfrac{\sqrt{2}}{2}$$

$$-\dfrac{\sqrt{2}}{2} \leq\sin2x\leq \dfrac{\sqrt{2}}{2}$$

$$0\leq2x\leq \dfrac{{\pi}}{4}$$,

$$\dfrac{3\pi}{4}\leq2x\leq {\dfrac{5\pi}{4}}$$,

$$\dfrac{7\pi}{4}\leq2x\leq {2\pi}$$,

$$\left[0;\dfrac{{\pi}}{8}\right] \cup \left[\dfrac{{3\pi}}{8};\dfrac{{5\pi}}{8}\right]\cup \left[\dfrac{{7\pi}}{8};{\pi}\right]$$