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: 
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 $(\cdots\cdots\cdots).$
To complement gimusi's fine answer, there are other cases when simple methods work:

1. $\cos f(x)=\cos g(x)$
2. $\sin f(x)=\sin g(x)$
3. $\tan f(x)=\tan g(x)$
4. $\sin f(x)=\cos g(x)$
5. $\cot f(x)=\tan g(x)$

where $f(x)$ and $g(x)$ are expressions involving the unknown $x$. 

Equation 1 has the solutions
$$
f(x)=g(x)+2k\pi
\qquad\text{or}\qquad
f(x)=-g(x)+2k\pi
$$

Equation 2 has the solutions
$$
f(x)=g(x)+2k\pi
\qquad\text{or}\qquad
f(x)=\pi-g(x)+2k\pi
$$

Equation 3 has the solutions
$$
f(x)=g(x)+k\pi
$$
(of course one has also to exclude values of $x$ that make $\tan f(x)$ or $\tan g(x)$ undefined).

Two angles have the same cosine if and only if the points on the unit circle they correspond to have the same $x$-coordinate; two angles have the same sine if and only if the points on the unit circle have the same $y$-coordinate. The $2k\pi$ or $k\pi$ term, with $k$ an integer, represents the periodicity.

What about an equation of the form $\sin f(x)=\cos g(x)$? We can recall that $\sin\alpha=\cos(\pi/2-\alpha)$, so we can reduce it to
$$
\cos\left(\frac{\pi}{2}-f(x)\right)=\cos g(x)
$$
which is type 1 above.

Similarly, $\cot f(x)=\tan g(x)$ can become
$$
\tan\left(\frac{\pi}{2}-f(x)\right)=\tan g(x)
$$
that is, type 3 above.