It is obvious for $a=0$. Now consider $a\ne0$.
We can find all the roots of form $x=e^{i\theta}$.
Divide the polynomial with $x^n$, we have equation
$2\cos(n\theta)+2a\sum_{k=1}^{n-1}\cos(k\theta)+a=0$.
Multiply by $\sin(\theta/2)$,
$\sin((n+\frac12)\theta)-\sin((n-\frac12)\theta)+a(\sin((n-\frac12)\theta)-\sin(\theta/2))+a\sin(\theta/2)=0$, or
$\sin((n+\frac12)\theta)-(a-1)\sin((n-\frac12)\theta)=0$, or
$\sin(n\theta)\cos(\theta/2)(2-a)+a\cos(n\theta)\sin(\theta/2)=0$, or
$\tan(n\theta)(1-\frac2a)=\tan(\frac{\theta}2)$.
We can easily see the equation has one root on $(\frac{2k-1}{2n}\pi,\frac{2k+1}{2n}\pi)$ for $k=1,\cdots,n-1$.
Paring with $-\theta$, we actually proved there are $n-1$ pairs conjugate roots to the polynomial of form $e^{i\theta}$.