Solving $\cos(3x) = \cos(2x)$

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 $(\cdots\cdots\cdots).$

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,

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