If $\cos 3x=\cos 2x$, then $3x=\pm 2x + 2\pi k$. Why the "$\pm$"? 
There is an equation I'm solving at the moment which involves $\arccos$. In the correction my teacher gave me, it seems after taking the $\arccos$ of an angle, you must take the positive and negative value of the angle plus a multiple of 2π:
Hence, solve:
 
$$ 4\cos^3(x) - 2\cos^2(x) - 3\cos(x) + 1 = 0, $$
For $0 ≤ x < \pi $
Solution:
$$\cos(3x) = 2\cos^2(x)-1$$
$$\cos(3x) = \cos(2x)$$
$$3x = ± 2x + 2\pi\times k$$
$$x=0, \  x=\frac{2}{5}\pi k$$
I was wondering why that is, and if there is an intuitive way of understanding this.
Thanks,
 A: Simply because $\cos(-x)=\cos(x)$ (as well as $\cos(x+2k\pi)=\cos(x)$).
A: $$ \cos x = \cos (-x) $$
So inverse function solution necessarily include its negative also.
$$ \cos x = \cos (\alpha) $$
$$ x= \pm \alpha  \pm 2 k \pi $$
In the present case
$$ 3x= \pm 2x + 2 k \pi$$
$$x=2 \pi k , \frac{2}{5}πk$$
Intuitive way is the graphical visual way, as an aid. For any inverse even function we have $\pm$ values necessarily, as the graph is symmetric to the x-axis.
EDIT1:
After posting the above, I realized giving the same symbol is part of the problem which came from not realizing that it can be factored. By setting each factor to zero, disambiguation is possible while recognizing two frequency waves are superimposed, and two symbols $(m,n)$ can or rather should be used.
$$ \cos 5x- \cos 3x =0,\quad -2 \sin 4 x \sin x =0 $$
$$x=2 \pi m , \frac{2}{5}π n$$
The $m$ wave roots are colored blue, and $n$ wave roots are green. Negative $x-$ axis graph (plotted in units of $\pi$) is not plotted because it is anyhow symmetrical as said above.
It can be seen why the roots labelled $(1,2,3,4,...)$ are double roots. In the interval required there is one real double root and two other real roots.

A: Notice, $$\cos(2k\pi+\theta)=\cos\theta$$$$ \ \ \cos(2k\pi-\theta)=\cos\theta$$
$$\cos(3x)=\cos(2x)\implies 3x=2k\pi\pm2x=\color{blue}{\pm2x+2k\pi}$$
Where, $k$ is any integer i.e. $k=0, \pm1, \pm2, \pm3, \ldots$
A: Use Prosthaphaeresis Formulas on $$\cos A=\cos B$$
$$2\sin\dfrac{B-A}2\sin\dfrac{B+A}2=0$$
If $\sin\dfrac{B-A}2=0, \dfrac{B-A}2=n\pi$ where $n$ is any integer
What if $\sin\dfrac{B+A}2=0?$
A: If you solve $$4x^3-2x^2-3x+1=0$$ then you get something like

with roots of $x=1,\frac{\sqrt{5}}{4}-\frac14,-\frac{\sqrt{5}}{4}-\frac14$
But now you want values which give those roots as their cosines. $0,\pm\frac{2\pi}5,\pm\frac{4\pi}5$ all do since $\cos$ is an even function but, since $\cos$ is also a periodic function, there are an infinite number of such roots, repeating every $2\pi$ and leading to the $+2n\pi$ in the solution.
$4\cos^3(x) - 2\cos^2(x) - 3\cos(x) + 1$ looks like the following and you can see the repeating solutions to your original equation

