Solutions of $\cos(ax)+\cos(bx)+\cos(cx)=0$

Question

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.

Answer 1

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.