How can I algebraically solve $\cos{3\theta} = \cos{2 \theta}, 0 \le \theta < \pi$ I know that $\theta = 0 $ is a solution by inspection; however, graphically, there are others. How can I determine these intersections without graphing both functions?
 A: Important trigonometric formulae.
$\cos{3\theta}- \cos{2 \theta}=-2\sin 5\theta/2 \sin \theta/2$
A: You use the (almost) primitive identity $$\cos \alpha=\cos\beta\iff \alpha=\pm\beta+2k\pi$$
A: First, we can rewrite both $\cos(3\theta)$ and $\cos(2\theta)$ as polynomials in $\cos(\theta)$ using various trigonometric identities:
$$\cos(2\theta)=2\cos^2(\theta)-1$$
$$\cos(3\theta)=4\cos^3(\theta)-2\cos(\theta)$$
Substituting these expressions into the original equation and moving all terms to one side gives an equation for the roots of a polynomial in $\cos(\theta)$:
$$4\cos^3\theta-2\cos^2\theta-3\cos\theta+1=0$$
Make the substitution $x=\cos\theta$ so that this equation looks a little bit more familiar:
$$4x^3-2x^2-3x+1=0$$
Solving this equation for $x$ and then solving for $\theta$ will give you the correct answer.
Note: The most obvious solution to this polynomial is $x=1$ which is also the most obvious solution to the original equation $\theta=0$, but you can factor out this solution to simply the cubic polynomial to a quadratic.
A: If $\cos(3 \theta) = \cos(2 \theta)$ then
$$3 \theta = \pm 2 \theta + 2 \pi k$$
with $k$ a integer.
A: From $\cos 3 \theta=\cos 2 \theta$ we have:
$$
3 \theta = \pm 2\theta +2k\pi \quad k \in \mathbb{Z}
$$
so the general solutions are:
$$
\theta=2k \pi \quad \lor \quad \theta= \frac{2k \pi}{5} \qquad k \in \mathbb{Z}
$$
and the solutions in $0\le \theta<\pi$ are $0, \frac{2\pi}{5},\frac{4\pi}{5}$
A: I posed the question, but I now realized the answer and decided to post it.
When one computes an $\arccos$ function in order to find an angle, there is another angle that also satisfies the original $\cos$ function; if $\theta$ is a solution to the $\cos$ function, then $2\pi - \theta$ is also a solution, although it doesn't fall within the domain of the $\arccos$ function.
Therefore:
$3\theta = 2\theta + 2k\pi,k \in \mathbb{Z}$ and $3\theta = 2\pi - 2\theta +2k\pi,k \in \mathbb{Z}$
So the solutions between $0$ and $\pi$ are:
$\theta = 2k\pi,k \in \mathbb{Z} \Rightarrow \theta = 0$
and
$5\theta = 2\pi + 2k\pi,k \in \mathbb{Z} \Rightarrow \theta = \dfrac{2\pi}{5} + \dfrac{2\pi}{5} k \Rightarrow \theta = \dfrac{2\pi}{5}, \dfrac{4\pi}{5}$
So, $\theta = 0, \dfrac{2\pi}{5}, \dfrac{4\pi}{5}$
