# Thread: countable set with all roots of trigonometric equations

1. ## countable set with all roots of trigonometric equations

Hey, can any one give me a hint of how to start this proof??

Let Q be the set of all that contains all roots of all trigonometric equations $P(sin x; cos x) = 0$ Prove that Q is countable.

2. So, what set does P range over? Depending on the answer, it may happen that the set of equations is countable.

3. Sorry, I omit that, but you are right, it is quite important. So P is a polynomial with integer coefficients

4. Since the set of such polynomials is countable, do you see the solution now?

5. Not really, I mean if you take as an assumption that every solution is countable, then it follows that the set of all solutions is countable. But is it that trivial to show (or are you assuming) that the solutions are countable?

For me the tricky part is to show that a solution to a trigonometric equation P(sinx,cosx)=0 is countable (where P(u,v) is a polynomial with integer coefficients). How you would show that?

6. You are right, this requires some thought. In fact, the set of roots is uncountable if the P(u, v) = 0.

If P is nontrivial, one way to show that the set of roots is at most countable is by proving that every root x of P(sin x, cos x) = 0 has a neighborhood that does not contain other roots. I am not sure now how to prove this for a two-variable polynomial. If we had just P(sin x) = 0, then we could make two steps. First, each root of P(x) has a root-free neighborhood; otherwise, P(x) would have infinitely many roots. Second, if P(sin x) = 0, then, by continuity of sin x, there is a neighborhood of x that is mapped into the root-free neighborhood of sin x found in step 1.