Solutions of $\cos(ax)+\cos(bx)+\cos(cx)=0$ These are  easy to solve:
$$\cos(ax)=0,\qquad \cos(ax)+\cos(bx)=0$$
Any insight into
$$\cos(ax)+\cos(bx)+\cos(cx)=0$$
Can't find any information on the web about this last one. I know solutions can be approximated via series but I'm interested in the exact cases or more information about triple exact solutions to identities like the last one.
 A: This isn't a complete answer, but I think it gets most of the way there.  
The first step would be to expand the cosines as follows:
$$
\cos(ax) = \cos([(a+c) - c]x)\\
\cos(bx) = \cos([(b+c) - c]x)\\
\cos(cx) = \cos(cx)
$$
With liberal use of trig angle formulae, $\cos(ax) + \cos(bx) + \cos(cx)$ becomes
$$
\cos((a+c)x)\cos(cx) + \sin((a+c)x)\sin(cx) + \\
\cos((b+c)x)\cos(cx) + \sin((b+c)x)\sin(cx) + \\
\cos(cx)
$$
Factoring will turn this into 
$$
\cos(cx)[\cos((a+c)x) + \cos((b+c)x) + 1] + \sin(cx)[\sin((a+c)x) + \sin((b+c)x)]
$$
To find out where this is zero, we can consider the case where the 2 halves are simultaneously zero.  Finding out where both halves are nonzero but cancel each other out is a headache that I don't want to deal with.  
So, we can consider the following 3 cases to get some of the solutions:
1) $\cos(cx) = 0$ and $\sin((a+c)x) + \sin((b+c)x) = 0$
or
2) $\sin(cx) = 0$ and $\cos((a+c)x) + \cos((b+c)x) = -1$
or
3) $\sin((a+c)x) + \sin((b+c)x) = 0$ and $\cos((a+c)x) + \cos((b+c)x) = 0$
At this point we're really just doing algebra.  For (1), I get 
$x = \frac{(\pi/2 + n\pi)}{c}$ and $\cos(ax) + \cos(bx) = 0$.
So
$$
\cos(\frac{a}{c}(\pi/2 + n\pi)) + \cos(\frac{b}{c}(\pi/2 + n\pi)) = 0
$$
This will only work for certain values of $a, b, c$.  Which values are left as an exercise to the reader.
Similarly, for (2), I get $x = n\pi/c$ and 
$$
\cos(cx)[\cos(ax) + \cos(bx)] = -1
$$
Again, only certain values of $a,b,c$ will result in solutions
For (3), I'll skip the details but I get that 
$$
x = \frac{4\pi}{3(a+c)} = \frac{5\pi}{3(b+c)}
$$
which also implies that an answer for this scenario exists only if 
$$
\frac{5}{4} = \frac{a+c}{b+c}.
$$
Expanding about $c$ was an arbitrary choice.  You can use the same method in either $a$ or $b$ to get the same solutions in terms of those variables. 
