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##