How to prove $\cos 36^{\circ} = (1+ \sqrt 5)/4$? Given $4 \cos^2 x -2\cos x -1 = 0$. 
Use this to show that $\cos 36^{\circ} = (1+ \sqrt 5)/4$, $\cos 72^{\circ} = (-1+\sqrt 5)/4$
Your help is greatly appreciated! Thanks

Hint: Look at the Quadratic Formula:
The solution to $ax^2+bx+c=0$ is $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

The equation is based on the fact that
$$
\cos(5x)=16\cos^5(x)-20\cos^3(x)+5\cos(x)
$$
and that $\cos(5\cdot36^\circ)=-1$ to get
$$
16\cos^5(36^\circ)-20\cos^3(36^\circ)+5\cos(36^\circ)+1=0
$$
Factoring yields
$$
(\cos(36^\circ)+1)(4\cos^2(36^\circ)-2\cos(36^\circ)-1)^2=0
$$
We know that $\cos(36^\circ)+1\ne0$; therefore,
$$
4\cos^2(36^\circ)-2\cos(36^\circ)-1=0
$$
Deciding between the two roots of this equation is a matter of looking at the signs of the roots.
For $\cos(72^\circ)$, use the identity $\cos(2x)=2\cos^2(x)-1$.
