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?

If $C=D=0$, then every $\theta\in\mathbb{R}$ is a solution, since the equation is just $0=0$.
If $D=iC\not=0$, we can divide by $C$ and the equation becomes
$$e^{i\sqrt\lambda\theta}=e^{i\sqrt\lambda(\theta+2m\pi)}$$
the solutions to which must satisfy $\sqrt\lambda\theta=\sqrt\lambda(\theta+2m\pi)+2n\pi$ for some $n\in\mathbb{Z}$, which means we must have $\sqrt\lambda m\in\mathbb{Z}$. This holds for all $m\in\mathbb{Z}$ if and only if $\lambda$ is a square integer.
If $D\not=iC$, then, choosing $\phi\in\mathbb{C}$ so that $\cos\phi=C/\sqrt{C^2+D^2}$ and $-\sin\phi=D/\sqrt{C^2+D^2}$, the equation becomes
$$\cos(\phi+\sqrt\lambda\theta)=\cos(\phi+\sqrt\lambda(\theta+2m\pi))$$
so we must have
$$\phi+\sqrt\lambda\theta=\pm(\phi+\sqrt\lambda(\theta+2m\pi))+2n\pi$$
for some $n\in\mathbb{Z}$. And once again, the only way to have this hold for all $m\in\mathbb{Z}$ is for $\lambda$ to be a sqare integer (in which case the solution with the plus sign is satisfied).
