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!

The equality $\cos(3x)=\cos(2x)$ is obviously true when $x=0$ and thus when $t=1.$ Therefore the polynomial
$$
4t^3-2t^2-3t +1
$$
has $t=1$ as one of its zeros. Consequently it can be factored:
$$
4t^3-2t^2-3t +1 = (t-1)(\cdots\cdots\cdots)
$$
The other zeros are those of a quadratic polynomial, written here as $(\cdots\cdots\cdots).$
