If y=2 cos(2 pi/7) then show y satisfy x^3+x^2-2x-1 =0?

Answer 1

Using complex numbers recall that
y = 2cos(z) = exp(iz) - exp(-iz)
Then the equation becomes
[exp(iz)-exp(-iz)]^3 + [exp(iz)-exp(-iz)]^2
-2[exp(iz)-exp(-iz)] - 1 = 0
i.e.,
Z^6+Z^5+Z^4+Z^3+Z^2+Z+1= 0
where
Z=exp(iz)
Note that
1/(1-Z) = Z^6+Z^5+Z^4+Z^3+Z^2+Z+1
+Z^7/(1-Z)
Thus our equation can be written as
(1-Z^7)/(1-Z)=0
with solution
Z^7 = 1
i.e.,
exp(7iz) = exp(2i*pi)
z = 2 pi/7
y = 2 cos(2 pi/7)

The other two roots of the cubic are
y = 2 cos(4 pi/7)
y = 2 cos(6 pi/7)
Do you see how to get them?