I want to solve this problem: $\cos(3x) = \cos(x)$ and I’m stuck.
I have tried to rewrite it to $4 \cos^3(x) - 3 \cos(x) = \cos (x)$ and then solve it. Add $3 \cos (x)$ to both sides and then divide by $4$. And then I have this: $\cos^3(x) = \cos (x)$ . I don’t know what so do now or even if on the right track. Please help, thanks.
You are on the right track. Start with
$$\cos 3x = \cos x$$
Substituting, $\cos 3x = 4 \cos^3x - 3\cos x$, you get
$$4\cos^3x-3\cos x = \cos x$$
Thus, $$\cos^3x-\cos x = 0$$
We can write this as $$\cos x(\cos^2x-1) = 0$$
By factoring the difference of two squares we get, $$\cos x(\cos x+1)(\cos x−1)=0$$
As we expected this equation has 3 factors and hence three solutions
Set each factor equal to 0 and solve using unit circle, here $n$ is a natural number.
$\cos x = 0$ which is satisfied for values, $x={±\frac{(2n-1)π}{2}}$
$\cos x+1=0$, thus $\cos x=−1$ which is satisfied for values, $x=nπ$
$\cos x−1=0$, thus $\cos x=1$ which is satisfied for values $x=0,2nπ$
As an alternative (and cleaner) solution, you can note that two angles $\theta, \theta'$ have the same cosine if and only if $\theta \equiv \theta' \pmod{2\pi}$ or $\theta \equiv -\theta' \pmod{2\pi}$. In your case, you thus have:
$$\begin{align} \cos(3x) = \cos(x) & \iff 3x \equiv x \pmod{2\pi}, \text{ or } 3x \equiv -x \pmod{2\pi} \\ & \iff 2x \equiv 0 \pmod{2\pi}, \text{ or } 4x \equiv 0 \pmod{2\pi} \\ & \iff x \equiv 0 \pmod{\pi}, \text{ or } x \equiv 0 \pmod{\pi/2} \\ & \iff x \equiv 0 \pmod{\pi/2} \end{align}$$ where, for the last equivalence, we use that if $x = k\pi$ for some integer $k$, then of course $x = 2k(\pi/2)$. So in the end, your solution set is: $$\mathcal{S} = \left\{ \frac{k \pi}{2} \mid k \in \mathbb{Z} \right\}.$$
