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!

By the definition of cosine function we have that

$$\cos \alpha = \cos \theta \iff \alpha = \theta +2k\pi \, \lor \, \alpha = -\theta +2k\pi \quad k\in \mathbb{Z}$$

and thus

$$\cos(3x) = \cos(2x)\iff 3x=2x+2k\pi \, \lor \, 3x=-2x+2k\pi \quad k\in \mathbb{Z}$$

that is

• $x=2k\pi$

• $x=\frac25 k\pi$

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 $(\cdots\cdots\cdots).$