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 $f(0)=a_n+...+a_1+a_0>0$.