General approaches to solve trigonometric polynomial equations

Question

Assume we are given an arbitrary trigonometric polynomial equation $$ f(\theta)=\sum^{n}_k(a_k cos(k\theta)+b_k sin(k\theta))=0, \ \ $$ where $\theta\in[0, 2\pi)$ and $a_k, b_k$ is real . I would like to ask:

1. Is there an approach that we can decide the number of real solutions given $\{a_k,b_k\}$?

2. Is there an approach that we can find out the analytical representation of real solutions?

Answer 1

This can be solved with the substitution $z = \tan\left(\dfrac{\theta}{2}\right)$

You can show that

$ \cos \theta = \dfrac{ 1 - z^2}{1 + z^2} $

$ \sin \theta = \dfrac{2z}{1 + z^2} $

Then you would have

$ \cos(n \theta) = \cos(\theta + (n-1) \theta) = \cos \theta \cos((n-1)\theta) - \sin(\theta) \sin( (n-1) \theta) ) $

and

$ \sin(n \theta) = \sin(\theta +(n-1) \theta) = \sin \theta \cos((n-1)\theta) + \cos \theta \sin((n-1) \theta) )$

So you can generate rational function of $z$ to express your function $f(\theta)$

Multiplying through by the least common denominator, generates a polynomial in $z$ which can be solved using well-established numerical methods.

Answer 2

You can rewrite $f$ as a (Laurent) polynomial of the complex exponential: $$\begin{split} f(\theta) & = \sum_{k=1}^n \left(a_k\frac{e^{ik\theta} + e^{-ik\theta}}{2}+b_k\frac{e^{ik\theta} - e^{-ik\theta}}{2i}\right)\\ &=\sum_{k=-n}^n c_ke^{ik\theta}\\ &= e^{-in\theta}\sum_{k=0}^{2n}c_{k-n}e^{ik\theta} \end{split}$$ In other words, $f(\theta)=0$ if and only if $P(e^{i\theta})=0$, where $P$ is the polynomial $$P(X)=\sum_{k=0}^{2n}c_{k-n}X^k$$ You thus need to look for the zeros of $P$ on the unit circle.

If $P$ is of degree $4$ or less, you can find a closed form for the roots of $P$, and check if there are any on the unit circle. If $P$ is of degree $5$ or above, you cannot expect, in general, a closed form for the roots.