Question: 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!

Answer: 
**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,

$$\sin\left(\frac{5x}{2}\right)=0 \text{ or } \sin\left(\frac{x}{2}\right)=0$$
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.

$$
x = \frac{2\pi}{5}k,\;\; \mbox{for}\;\; k=0,1,2,\cdots
$$