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

Using $\cos5\theta = 16\cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta$;
$\cos 180 = 16\cos^5 36 - 20 \cos^3 36 + 5 \cos 36$
Let $\cos36 = x$:
$-1 = 16x^5 -20x^3 +5x$
Its solutions are ${-1,\frac{1- \sqrt{5}}{4}},\frac{1+ \sqrt{5}}{4}$
And as $\cos36 = x$, $\cos 36$ must be equal to one of them.
$\cos 36$ must be equal to $\frac{1+\sqrt{5}}{4}$, as $-1$ and $\frac{1- \sqrt{5}}{4}$ are negative.
