Let $\theta=\frac{2\pi}{5}$. Show $2\cos(2\theta)+2\cos(\theta)+1=0$. I have been at this for a while. Any ideas?

I think this is a simple way. By using De Moivre’s formula, we obtain the fifth time angle formula:
\begin{equation}
\cos(5\theta)=16\cos^5\theta-20\cos^3\theta+5\cos\theta
\end{equation}
Set $\theta=2\pi/5$, we have:
\begin{equation}
\cos(2\pi)=16\cos^5\theta-20\cos^3\theta+5\cos\theta\\
16\cos^5\theta-20\cos^3\theta+5\cos\theta-1=0
\end{equation}
By using factorization, we have:
\begin{equation}
(\cos\theta-1)(4\cos^2\theta+2\cos\theta-1)^2=0
\end{equation}
Obviously, $\cos\theta-1\neq0$, then the equality have been prove.
