$|2\sin^2 x-1|>\cos x$: graphic and analytical method resolution

calculatoreducationmath-softwaresoft-questiontrigonometry

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:

enter image description here

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.?

Best Answer

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.