Find the roots of $z^4-3z^2+1=0$ in polar form. Question :
Prove that the solutions of $z^4-3z^2+1=0$ are given by :
$$z=2\cos{36^\circ},2\cos{72^\circ},2\cos{216^\circ},2\cos{252^\circ}$$
My work :
First of all, i want ro find the roots with quadratic formula
$\begin{align}
&(z^2)^2-3z^2+1=0\\
&z^2=\dfrac{3\pm \sqrt{5}}{2}\\
&z_{1,2}=\pm\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\\
&z_{3,4}=\pm\dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}\\
\end{align}$
And i'm stuck. I don't know how to transform this complicated roots into polar form. Bcz, i'm not sure that the modulus for each number is exactly $2$?
Btw, i've found and read some possible duplicates
Here's one of the links :
Roots of $z^4 - 3z^2 + 1 = 0$.
But it seems doesn't answer my question...
Please give me a clear hint or another way to solve this without quadratic formula or something else.

Note that 
$$\cos36 = -\cos216 = \frac{1+\sqrt5}{4}$$
$$\cos72 = -\cos252 = \frac{1-\sqrt5}{4}$$
Then,
$$(x-2\cos36)(x-2\cos216) = x^2 - \frac{3+\sqrt5}{2}=0$$
$$(x-2\cos72)(x-2\cos252) = x^2 - \frac{3-\sqrt5}{2}=0$$
Together,
$$(x-2\cos36)(x-2\cos216)(x-2\cos72)(x-2\cos252)=x^4-3z^2+1=0$$
