Suppose $f$ is a real trigonometric polynomial of degree $N$ with constant term $0$. What lower bounds can we place on the measure $\mu$ of the set $\{ t \in S^1 : f(t) \geq 0 \}$, independent of the coefficients of the polynomial?
Using Parseval to put a lower bound on $\max |f|$, and putting a crude upper bound on $|f'|$, I can get a lower bound on $\mu$ of order $N^{-3/2}$. But I'm guessing that one can probably do significantly better.
Just to fix notation, I'm putting the factors of $2 \pi$ inside the trig functions, so that $f$ has period $1$ and $S^1$ has length $1$.
Added in response to comment:
Let $C$ be the sum of the absolute values of the coefficients of $f$, and let $D$ be the sum of the squares of the coefficients. Let $M = \max |f|$. By Parseval we have $M \geq \alpha D^{1/2}$ for some constant $\alpha \in \mathbb{R}_+$. Then by the QM-AM inequality we get $M \geq \alpha C N^{-1/2}$. Trivially we have $|f'| \leq \pi C N$.
Suppose $\max f = M$, and that this max is attained at $0$. The interval about $0$ on which $f \geq 0$ then has length at least
$$2 \frac{\max f}{ \max |f'|}$$
(assuming $f$ isn't identically zero), which is at least
$$\frac{2 \alpha}{\pi} N^{-3/2}$$
Otherwise, $\max f < M$ and the above calculation shows that the graph of $f$ has a triangle of area at least
$$\frac{M \alpha}{\pi} N^{-3/2}$$
lying below the $t$-axis. Since $f$ integrates to zero, the integral of the non-negative part of $f$ must cancel out this contribution. Since $f < M$ we see that $f$ must be non-negative on a set of length at least
$$l := \frac{\alpha}{\pi} N^{-3/2}$$
So in either case we get a lower bound of $l$.
I have the same upper bound as Thursday.
An easy lower bound is $\frac1{N+1}$: it follows from the identity $$ f(x)+f\left(x+\frac1{N+1}\right)+f\left(x+\frac2{N+1}\right)+\ldots+f\left(x+\frac{N}{N+1}\right)=0. $$