Solving $\cos(3x) = \cos(2x)$ I'm struggling with solving given trigonometric equation
$$\cos(3x) = \cos(2x)$$
Let's take a look at the trigonometric identities we can use:
$$\cos(2x) = 2\cos^2-1$$
and 
$$\cos(3x) = 4\cos^3(x) -3\cos(x)$$
Plugging into the equation and we have that
$$4\cos^3(x) -3\cos(x) = 2\cos^2(x)-1$$
$$4\cos^3(x) -3\cos(x) - 2\cos^2(x)+1= 0$$
Recalling $t = \cos (x)$, 
$$4t^3-2t^2-3t +1 = 0$$
Which is a cubic equation. Your sincerely helps will be appreciated. 
Regards!

Calling 
$$
\cos x = \frac{e^{ix}+e^{-ix}}{2}
$$
we have
$$
e^{3ix}+e^{-3i x} = e^{2ix}+e^{-2i x} 
$$
or calling $z = e^{ix}$
$$
z^6+1 = z^5+z\to z^6-z^5-z+1 = (z-1)^2(z^4+z^3+z^2+z+1) = (z^5-1)(z-1) = 0
$$
so the solutions are obvious.
$$
x = \frac{2\pi}{5}k,\;\; \mbox{for}\;\; k=0,1,2,\cdots
$$
