Find the number of roots of $f(x)$ in the interval $0\leq x\leq 2\pi$

Question

If $$f(x)=a_{0}+a_{1}\cos(x)+a_{2}\cos(2x)+\cdots+a_{n}\cos(nx)$$ where $a_{0}, a_{1},\ldots,a_{n}$ are non-zero real numbers and $$a_{n}>\lvert a_{0}\rvert+\lvert a_{1}\rvert+\cdots \lvert a_{n-1}\rvert$$ then the number of roots of $f(x)$ in the interval $0\le x\le 2\pi$ is:

The choices/options are:Atmost $n$ , more than $n$ but less than $2n$ , at least $n$ , zero

My try:

$$\lvert f(x)\rvert \le \lvert a_{0}\rvert+\lvert a_{1}\rvert+\cdots \lvert a_{n-1} \rvert+\lvert a_{n} \rvert$$ $$\implies \lvert f(x)\rvert \lt a_{n} +\lvert a_{n} \rvert$$ $$\implies \lvert f(x)\rvert \lt 2a_{n}$$ I am unable to proceed further...Thanks for any answers!

Answer 1

Hint: when $\cos(x) = \pm 1$ what can you say about $f(x)$?

Answer 2

Here's a more extended hint along the lines of my comment: at the places where $\cos(nx) = \pm 1$, the last term is bigger in absolute value than the sum of all the lower terms - can you see why?

That means that the sign of the last term is the same as the sign of the function at those places. So find those places where $\cos (nx) = \pm 1$ and consider adjacent ones: at one place, the last term is positive and so is the function and at the next point the last term is negative and so is the function. So what can you say about the behavior of the function (and in particular its zeros) in between?

Then consider all such adjacent intervals between 0 and $2\pi$ and count.