How to determine whether all solutions to $\sin(ax) + \sin(bx) + \sin(cx)=0$ in are rational multiples of $\pi$
I was messing around on Desmos trying to create trigonometry problems when I came across the following:
For what positive integers $a,b,c$ is it true that all possible roots of $$\sin(ax)+\sin(bx)+\sin(cx)=0$$ are rational multiples of $\pi?$
By inspection I found some triples $(1,2,3), (1,3,4), (1,3,5), (2,3,4), (3,5,7)$ but I do not see any pattern. I looked into Chebyshev polynomials but that seems extremely ugly. How would I go about determining
- whether there are infinitely many triples $(a,b,c)$
- what is the "criteria" for such a triple?
Partial answer.
Assume that $a<b<c.$ If $a,b,c$ is an arithmetic progression, then $b=\frac{a+c}2,$ and $\frac{c-a}2=b-a,$ so $$\sin (ax)+\sin(cx)=2\sin(bx)\cos((b-a)x).$$
So in this case, the equation is equivalent to $$0=\sin(bx)(1+2\cos((b-a)x))$$ and you have that all roots are rational multiples of $\pi.$
Not sure if this is necessary, but it is sufficient.
More generally, let $f(x)$ be a linear combination over the rationals of sines of positive integer multiples of $x$. Taking $w = e^{ix}$, so $\sin(ax) = (w^a - w^{-a})/(2i)$, this can be written as $P(w)/(i w^n)$ where $P$ is a polynomial with integer coefficients. All (real or complex) zeros of your $f$ are rational multiples of $\pi$ if and only if $P$ is a constant times a product of cyclotomic polynomials; all real zeros are rational multiples of $\pi$ if and only if $P$ is a constant times a product of cyclotomic polynomials times a polynomial with no roots on the unit circle.