Trigonometry : verify that $\cos \frac{A}{2} +\cos 2A = 0$ for $A=\frac{2\pi}{5}$

Trigonometry : verify that $\cos \frac{A}{2} +\cos 2A = 0$ for
$A=\frac{2\pi}{5}$

Question in trigonometry: verify that $\cos\frac{A}{2} + \cos 2A = 0$.
Let $A = \frac{2\pi}{5}$.
Verify that $\cos \frac{A}{2} + \cos 2A = 0$.
Show that $x=\cos A$ satisifies the equation $x=2(4x^4-4x^2+1)-1$.
Factor $8x^4-8x^2-x+1$ over $\mathbb Z$, and deduce that $\cos A$ is a
zero of a quadratic polynomial over $\mathbb Z$.
Determine $\cos A.$
My solution is as follows : We know cos (A/2)= ¡î[(1+ cosA)/2)] and cos 2A
= 2cos©÷A-1
cos2A= 2cos©÷A-1
It is sufficient if we prove that cos2A= - cos(A/2) for the first question.
¢¡2cos©÷A-1 =¡¾ ¡î[(1+ cosA)/2)]
squaring on both sides and simplifying: 8cos4A + 2 - 8 cos©÷A= 1 + cosA
subsitituting cos A with x :
8x^4- 8x©÷+2-1 = x
2(4x^4-4x©÷+1)-1=x
I believe that if I do the same thing back wards, I will get the answer
for question (2)
I have factorized the equation and ended up with this: (x-1)[8x©ø+8x-1]
Now, the third question asks me to deduce that cosA is a zero of quadratic
polynomial, but I have only linear and cubic factors, how do I proceed?
Regarding the first question, I¡¯m clueless, I have simplified it as much
as possible, but its not going anywhere.