Given that

$e^{ix}=\cos x+i\sin x$

imaginary roots always occurs in pairs

$e^{-ix}$ is also solution of

$z^n+p_1{z}^{n-1}+p_2{z}^{n-2}+\dots \dots +p_n=0$

i.e.,

$p_n{\left(e^{ix}\right)}^n+p_{n-1}{\left(e^{ix}\right)}^{n-1}+p_{n-2}{\left(e^{ix}\right)}^{n-2}+.....p_1{e}^{ix}+1 = 0$

$\Rightarrow p_n\left(\cos nx+i\sin nx\right)+p_{n-1}\left(\cos \left(n-1\right)x+i\sin \left(n-1\right)x\right)+p_{n-2}\left(\cos \left(n-2\right)x+i\sin \left(n-2\right)x\right)+......p_1\left(\cos x+i\sin x\right)+1=0$

By comparing real and imaginary roots

$p_n\sin nx+p_{n-1}\sin \left(n-1\right)x+\dots +p_1\sin x+1=0.$