Roots of the polynomial involving trigonometric function

Question

Given $$f(x)=a_0+a_1\cos x+a_2\cos2x+\cdots+a_n\cos nx$$ where $a_0,a_1,a_2,\cdots,a_n$ are non-zero reals such that $a_n>|a_0|+|a_1|+\cdots+|a_n-1|$ then how many roots of $f(x)=0$ are there in $0\le{x}\le{2\pi}$?
I have tried to check the conditions for lower power equations. For $a_0+a_1\cos x=0$ there are two roots and for $a_0+a_1\cos x+a_2\cos 2x=0$ I am getting four roots. After this I tried to use induction on the equation but could not find any possible way to solve the problem. So how can we approach this type of problem?

Answer 1

Part 1: $f(x)$ has at least $2n$ roots in $[0,2\pi]$.

Just evaluate $f(x)$ at $x = \frac{k\pi}{n}$ for $k=0,\ldots,2n.$ Then $$ f\left(\frac{k\pi}{n}\right) = \sum_{m=0}^{n-1} a_m\cos\left(\frac{km\pi}{n}\right) + a_n\cos\left(k\pi\right) = \sum_{m=0}^{n-1} a_m\cos\left(\frac{km\pi}{n}\right) + a_n (-1)^k $$ Because of $a_n > \sum_{m=0}^{n-1}|a_m|,$ the sign of $f(k\pi/n)$ is determined by $(-1)^k.$ This means that the sign of $f(x)$ changes at least $2n$ times between $0$ and $2\pi,$ which in turn means that there must be at least $2n$ roots.

Part 2: $f(x)$ has at most $2n$ roots in $[0,2\pi]$.

Note that $x=0$, $x=\pi$ and $x=2\pi$ cannot be roots, because $|\cos nx|=1$ in that case and $a_n\cos nx$ consequently dominates the other terms of the sum.

We set $x=\arccos u$ and use the ordinary convention that the result of $\arccos$ is in $[0,\pi]$. Then $$ f(\arccos u) = \sum_{m=0}^{n} a_m\cos\left(m\arccos u\right) =\sum_{m=0}^{n} a_m T_m(u) =: g_1(u) $$ with the Chebyshev polynomials of the first kind $T_m(u).$ Therefore, $g_1(u)$ is a polynomial of degree $n$ and can have at most $n$ roots which in turn means that $f(x)$ can have at most $n$ roots in $(0,\pi).$

Likewise, you can define $x=\pi+\arccos v$ and you get $$ f(\pi+\arccos v) = \sum_{m=0}^{n} a_m\cos\left(m\pi+m\arccos v\right) =\sum_{m=0}^{n} (-1)^m a_m T_m(v) =: g_2(v) $$ $g_2(v)$ is also a polynomial of degree $n$ and can have at most $n$ roots which in turn means that $f(x)$ can have at most $n$ roots in $(\pi,2\pi).$

Put all this together, and you get that $f$ has exactly $n$ roots in $(0,\pi)$ and exactly $n$ roots in $(\pi,2\pi)$.