$|2\sin^2 x-1|>\cos x$: graphic and analytical method resolution I have this inequality
$$|2\sin^2 x-1|>\cos x \iff |\cos(2x)|>\cos x \tag 1$$
or equivalently to
$$|1-2\cos^2x|>\cos x \tag 2$$
To solve the $(2)$ is trivial by taking it to an absolute value inequality of the second degree, but it would waste a lot of time, at least in my opinion. I did this exercise this morning to my 17 year old high school students using the graphical method with the $(1)$.
First I hand drew everything keeping in mind that the principal period $T=\pi$ for $\cos(2x)$. My graph it was correct and it is similar to this done with Geogebra:

I have said that all the solutions are just that between the $y$-value point of $A$ and the $y$-value point of $B$. Being $A\equiv(\ldots; 0.5)$ and $B\equiv(\ldots; 0.5)$ we have that the angles $x$ are, $\pi/3$ and $2\pi-\pi/3$.

For my students, to speed up the calculations, without the use of Geogebra how is it possible to find the coordinates of points $A$ and $B$ using the scientific calculator or some simple method other than numerical methods, derivatives, series expansion, etc.?

 A: Notice that the intersections are solutions of
$$\cos(2x)=\pm\cos(x),$$
hence
$$2x=\pm x+2k\pi$$ or $$2x=\pm x+(2k+1)\pi.$$
Finally,
$$x=\frac{k\pi}3.$$
The rest is a matter of sign discussion.
A: I include my answer...
Remark: In $\Bbb R$,  $|a|=b\iff (b\ge0\wedge(a=\pm b))$.
For this remark we have:
$$\cos(2x)=|\cos(x)|\iff(\cos(2x)\ge0)\wedge((\cos x=\cos(2x))\vee(\cos x=-\cos(2x)).$$

*

*The equation $\cos(2x)=+ \cos (x) \iff \cos(2x)-\cos (x)=0$ not give me any solution for to find $A$ and $B$, but

$$\cos(2x)+\cos (x)=2\cos\left(\frac{3x}{2}\right)\cos\left(\frac{x}{2}\right)=0.$$
For the
$$\cos \left(\frac{3x}{2}\right)=0$$  I will have
$$x=\frac{\pi}3+\frac{4}3k\pi, \quad x=\pi+\frac{4}3k\pi \tag 1$$
For the $$\cos \left(\frac{x}{2}\right)=0$$ I will have
$$x=3\pi+4k\pi, \quad x=\pi+4k\pi \tag 2$$
Merge overlapping the intervals $(1)$ and $(2)$ I will have
$$x=\frac{\pi }{3}+\frac{4\pi k}{3},\:x=\pi +\frac{4\pi k}{3}$$
and the accepted solutions are only
$$x=\frac{\pi }{3}+\frac{4\pi k}{3}$$
