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!

Hint
You can use sum-product equivalence. Which is:
$$\cos(A)-\cos(B)=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$
so,
$$\cos(3x)-\cos(2x)=0\to-2\sin\left(\frac{3x+2x}{2}\right)\sin\left(\frac{3x-2x}{2}\right)=0$$
$$\sin\left(\frac{5x}{2}\right)\sin\left(\frac{x}{2}\right)=0$$
so,
$$\sin\left(\frac{5x}{2}\right)=0 \text{ or } \sin\left(\frac{x}{2}\right)=0$$
