How do you find the exact values of cos 2pi/5?

##cos(2pi/5)=(-1+sqrt(5))/4##
Here the most elegant solution I found in:
http://math.stackexchange.com/questions/7695/how-to-prove-cos-frac2-pi-5-frac-1-sqrt54
##cos(4pi/5)=cos(2pi-4pi/5)=cos(6pi/5)##
So if ## x=2pi/5##:
##cos(2x)=cos(3x)##
Replacing the cos(2x) and cos(3x) by their general formulae:
##color(red)(cos(2x)=2cos^2x-1 and cos(3x)=4cos^3x-3cosx)##
we get:
##2cos^2x-1=4cos^3x-3cosx##
Replacing ##cosx## by ##y##:
##4y^3-2y^2-3y-1=0##
##(y-1)(4y^2+2y-1)=0##
We know that ##y!=1## so we have to solve the quadratic part:
##y=(-2+-sqrt(2^2-4*4*(-1)))/(2*4)##
##y=(-2+-sqrt(20))/8##
since ##y>0## ##y=cos(2pi/5)=(-1+sqrt(5))/4##