Using the identity pointed out by J.M., we can transform the equation into
$$\cos(\mu-\nu)-\cos(\mu+\nu) = C(\cos(\mu-\nu)+\cos(\mu+\nu)) \tag1$$
or better yet,
$$(1-C)\cos(\mu-\nu) = (1+C) \cos(\mu+\nu) \tag2$$
Here both $\mu-\nu$ and $\mu+\nu$ are linear functions of $t$, which, as I infer from a comment by the OP, means linear and not affine. Assuming neither $\mu-\nu$ nor $\mu+\nu$ happen to be identically zero (which is an easy case), we can make (2) more compact:
$$\cos x = A\cos bx \tag3$$
where $A=(1+C)/(1-C)$ and $b=(\mu+\nu)/(\mu-\nu)$, which is a nonzero constant. There are two cases:
$b$ is rational, $b=m/n$. Then both sides of (3) are periodic functions with period $2\pi n$. Let a numerical routine find all roots of (3) on $[0,2\pi n]$, and write down all of them by periodicity. If both $m$ and $n$ are very small, an explicit solution may exist, but otherwise not.
$b$ is irrational. The function $\cos x-A\cos bx$ is not periodic, but it's quasiperiodic. This means that the roots of (3) come arbitrarily close to forming a periodic pattern, but they never do. Here is the graph of $\cos x-\pi \cos ex$, for example:
All I can add is an easy observation: if $A\ll 1$ or $A\gg 1$, the roots of (3) are close to the zeroes of the wave with greater magnitude.
I wonder if (3) has a name...