Solving $C\cos(\sqrt\lambda\theta)+D\sin(\sqrt\lambda\theta)=C\cos(\sqrt\lambda(\theta+2m\pi)) + D\sin(\sqrt\lambda (\theta + 2m\pi))$ 
I want to solve
$$C\cos(\sqrt\lambda \theta) + D\sin(\sqrt\lambda \theta) = C\cos(\sqrt\lambda (\theta + 2m\pi)) + D\sin(\sqrt\lambda (\theta + 2m\pi))$$
The solution must be valid for all $\theta$ in $\mathbb{R}$ and all $m$ in $\mathbb{Z}$, but $C$, $D$, and $\lambda$ are to be determined and can be in $\mathbb{C}$.

The solutions I've found by guessing are $(C\ $arbitrary$, D\ $arbitrary$, \lambda = n^2)$, where $n$ is any integer, and $(C = 0, D = 0, \lambda\ $arbitrary$)$.
Is there some algebra I can do to show that these are the only solutions, or find the rest of the solutions to this equation?

Use the formula $\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$ and  $\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$. When you substitute this into your expression, on the left-hand side, you would get $ C(\frac{e^{i\sqrt{\lambda}\theta} + e^{-i\sqrt{\lambda}\theta}}{2}) + D(\frac{e^{i\sqrt{\lambda}\theta} - e^{-i\sqrt{\lambda}\theta}}{2i})$. On the right-hand side, you would end up with
$C(\frac{e^{i\sqrt{\lambda}\theta} e^{2\pi mi\sqrt{\lambda}} + e^{-i\sqrt{\lambda}\theta}e^{2\pi mi\sqrt{\lambda}}}{2}) + D(\frac{e^{i\sqrt{\lambda}\theta}e^{2\pi mi\sqrt{\lambda}} - e^{-i\sqrt{\lambda}\theta}e^{2\pi mi\sqrt{\lambda}}}{2i})$. Notice how the right-hand-side is similar to the left-hand side, except of the $e^{2\pi mi\sqrt{\lambda}}$ term. If we set this equal to 1 and solve for $\lambda$, we would see that $\sqrt{\lambda}$ would have to be an integer and therefore making $\lambda = n^2$
This explains the one solution for any C or D value. If you want to find any arbitrary C or D value, I would start by using Euler's formula to turn $C\cos{\sqrt{\lambda}\theta} + D\sin{\sqrt{\lambda}\theta}$ into one function,  and to do the same with the right-hand side.
