Check polynomial $p_n(x)$ is always divisible by $x^2 - 2x\cos (t) +1$ For any real $t$ and $n \in\mathbb{N}$, I want to show that $p_n(x)=x^{n+1}\cos [(n-1)t] - x^n\cos (nt) - x\cos (t) +1$ is divisible by $q(x) = x^2 - 2x\cos (t) +1$. As a hint I have: complex numbers.
I don't know how to attack the problem using complex numbers. In the case $n=1$, $p_1(x)=q(x)$. Also I compute and check by hand the case $n=2$. Then I start thinking about induction, but the idea failed.
Any idea how this can be solved via complex numbers? Thank you in advance.

This is equivalent to $z^{\pm1}$ being roots, with $z:=e^{it}$. We only need check $\pm=+$, since $p_n$ has real coefficients, so its nonreal roots are in conjugate pairs. So just simplify$$p_n(z)=z^{n+1}(z^{n-1}+z^{1-n})/2-z^n(z^n+z^{-n})/2-z(z+z^{-1})/2+1.$$
