Students are asked to solve trigonometric equations shortly after learning what sine and cosine are. By some combination of persistence and luck they may be able to find a solution. After proudly presenting the solution to a teacher, the teacher may ask “Is that the only solution?” A candid student would respond by saying “How should I know?!”. The omniscient teacher, having seen the solution, declares that there is another.

Trigonometric equations pop up fairly often in applications, but thoroughly solving such equations requires algebraic techniques that are not typically part of the undergraduate curriculum.

This post will present a terse outline of a systematic way to approach trigonometric equations that are polynomials in sines and cosines of integer multiples of an angle θ.

First, get rid of all the multiples of θ by using multiple angle identities.

By repeatedly applying these identities we can reduce terms involving integer multiples of θ to terms involving only sums and products of sin(θ) and cos(θ).

Next, given a trigonometric equation of the form

where *P* is a polynomial in two variables, we turn it into a system of two polynomial equations

by introducing *s* = sin(θ) and *c* = cos(θ).

Now we have a purely algebraic system of two polynomials in two variables. A system of two polynomials in *one* variable have a common solution if and only if their resultant is zero. We can think of a polynomial in two variables as a polynomial in one variable whose coefficients are polynomials in the other variable.

If we think of our polynomial in *s* and *c* as a polynomials *s* with coefficients given by polynomials in *c*, then the condition that the resultant equals zero gives a polynomial equation in *c* that solutions must satisfy.

In short, we reduce a trigonometric equation to a system of two polynomial equations in two unknowns, and then to a single polynomial equation in one unknown. In general the final polynomial will have high degree, and so finding its roots will take some work, but it’s a routine problem.

I think you can reduce the system to f(c) = s g(c) and s^2=1-c^2, and that squaring the first and substituting for s^2 gives you the same equation as the resultant. It might be easier to explain to high school students.