How to prove $a_0 + a_1 \cos \theta + a_2 \cos 2\theta + \cdots + a_n \cos n \theta$ has $2n$ different zeros, $\theta \in (0,2\pi)$. $0 < a_0 < a_1 < \cdots < a_n$.
Prove that $a_0 + a_1 \cos \theta + a_2 \cos 2\theta + \cdots + a_n \cos n \theta$ has $2n$ different zeros, $\theta \in (0,2\pi)$.
[Hint: First prove that $P_n(z)=a_o+a_1z+a_2z^2+\cdots+a_nz^n$ has $n$ zeros in unit ball $B(0,1)$.]
This is an assignment I copied from my textbook. It's in the section "The Argument Principle & Rouche Theorem".
Though I followed this hint, I still can't see how this would imply the desired conclusion. Help needed. 
 A: Direct solution via intermediate value theorem
Multiply with the positive (on $(0,2\pi)$) factor $2\sin\fracθ2$ to get the equivalent equation
\begin{multline}
0=g(θ)=(a_0-a_1)\sin\tfrac12θ+(a_1-a_2)\sin\tfrac32θ+…\\…+(a_{n-1}-a_n)\sin((n-\tfrac12)θ)+a_n\sin((n+\tfrac12)θ)
\end{multline}
At the extremal points of $\sin((n+\tfrac12)θ)$, which are $$θ_k=\frac{2k+1}{2n+1}\pi, ~~~ k=0,1,...,2n,$$ the value $g(θ_k)$ has the same sign $(-1)^k$ as its last term. This is because it dominates the sum of the other terms, as $$a_n>a_n-a_0=\sum_{k=0}^{n-1}|a_k-a_{k+1}|.$$ This is evidence for $2n$ sign alternations in the function value and thus at least $2n$ real roots inside the given interval $(0,2\pi)$.
Solution according to the hint
Per the hint, try to locate all the roots inside the unit circle. As the coefficient sequence is separated by inequalities, one can modify the coefficients slightly without destroying this defining property. Then multiplying with a linear factor with a root at $1$ gives
$$
q(z)=(z-1)p(rz)=r^{n}a_nz^{n+1}+(r^{n-1}a_{n-1}-r^{n}a_n)z^n+...+(a_0-ra_1)z-a_0.
$$
The roots of $q(z)$ are contained in a circle of radius
$$
R=\max(1,r^{-n}|a_n|^{-1}(|r^{n-1}a_{n-1}-r^{n}a_n|+...+|a_0-ra_1|+|a_0|))=1.
$$
This bound is valid as long as $ra_{k+1}\ge a_k$, $k=0,..,n-1$. There is some $r<1$ that satisfies this finite number of inequalities.
So if $z$ is a root of $p$, then $z/r$ is a root of $q$, thus $|z/r|\le 1$, $|z|<r$. All roots of $p(z)$ are well inside the unit circle.
The path $p(e^{iθ})$, $θ\in[0,2\pi)$ of the image of one rotation along the unit circle has winding number $n$ around zero. Which means it crosses the positive real half axis at least $n$ times and also the negative real half-axis at least $n$ times. These crossing points are also roots for $f(θ)=Re(p(e^{iθ}))$, the function under consideration. Thus
$$
f(θ)=a_0 + a_1 \cos \theta + a_2 \cos 2\theta + \cdots + a_n \cos n \theta
$$
has at least $2n$ roots in $(0,2\pi)$. Note that $f(0)=a_n+...+a_1+a_0>0$.
A: The problem you have provided does not appear to be a simple application of Rouche's theorem or equivalent. I would be curious what the author's thoughts were for a proof. In particular, I believe the problem is equivalent to the following non-trivial theorem.
Theorem. If $0<a_0<a_1<\cdots<a_n$ and $p(z)=a_0+\cdots+a_nz^n$, then $$q(z):=z^{-n}p(z^{-1})+z^np(z)$$ has only simple zeros located on the unit circle.
The equivalency can be easily verified. Indeed, if we define $$c(z):=a_0+a_1\cos(z)+a_2\cos(2z)+⋯+a_n\cos(nz)\ \ \ ,$$ then the statement "$c(z)$ has $2n$ simple zeros in the interval $(0,2\pi)$" is equivalent to the statement that "$q(z)$ has $2n$ simple zeros on the unit circle". This is because $\cos(z)=(e^{iz}+e^{-iz})/2$, hence $$c(z)=e^{-inz}/2*q(e^{iz})\ \ \ .$$ Thus $c(z)$ has a zero in $(0,2\pi)$ if and only if $q(z)$ has a zero on the unit circle.
I would gladly welcome a simple proof of the above theorem with an application of Rouche's theorem or similar, but this appears to not be the case. In fact, if you generalize $p(z)$ to polynomials with zeros in the unit disc and throw in a rotation coefficient of $|a|=1$, the above theorem is precisely the complete characterization of all polynomials with zeros on the unit circle, something that was only established in 1995. See the first answer here To prove this complex polynomial has all zeros on unit circle, Chen (J. of Math. Anal. and Appl. vol 190, 714-724 (1995)).
