Solve $x^4-3x^2+1=0$ in terms of cosine. I put the equation in the form of a quadratic:
$(x^2)^2-3x^2+1=0$
Then using the quadratic formula,
$x^2=\frac{3\pm\sqrt{9-4}}{2}$
$x^2=\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$
$x=\pm\frac{1+\sqrt{5}}{2}$ and $\pm\frac{1-\sqrt{5}}{2}$
So there are four roots as expected given the equation is a quartic. But I really don't know how to put the answers in terms of cosine. Any hints?
 A: $\cos(4 t) = 8 \cos(t)^4 - 8 \cos(t)^2 + 1$, so if $x = \sqrt{3} \cos(t)$,
 $$x^4 - 3 x^2 = 9 (\cos(t)^4 - \cos(t)^2) = \dfrac{9}{8} (\cos(4t) - 1)$$ 
Thus to solve $x^4 - 3 x^2 + c = 0$, take $t = \arccos(1 - 8c/9)/4$ and 
$$x = \sqrt{3} \cos(t) = \sqrt{3} \cos \left( \dfrac{1}{4} \arccos\left(1 - \dfrac{8c}{9}\right)\right)$$
If you want this solution to be real, you need $1 \ge 1 - 8c/9 \ge -1$, i.e.
$0 \le c \le 9/4$.
A: let $\tau = \dfrac{1 + \sqrt 5}{2}$  be the golden ratio which is the positive solution of the quadratic equation $x^2 - x - 1=0$
we also have all powers of $\tau$ as linear combination of $1, \tau$ for example, $ \dfrac{1}{\tau} = \dfrac{\sqrt 5 -1}{2}, \tau^2 = \tau + 1 =\dfrac{3+\sqrt 5}{2}, \dfrac{1}{\tau ^2} = 2 - \tau = \dfrac{3 - \sqrt 5}{2}, \cdots$
from the regular pentagon you get $\cos 36^\circ = \tau^2/2, \cos 72^\circ = \dfrac{1}{2\tau}$  the quadratic equation that has $\tau^2, 1/\tau^2$ for roots is $(x - \tau^2)(x - 1/\tau^2) = x^2 - 3x + 1$
so the roots of the quartic equation $x^4 - 3x^2 + 1 = 0$ are $\pm \tau, \pm \dfrac{1}{\tau}.$
we can express the two roots $\dfrac{1}{\tau} = 2 \cos 72^\circ, \tau^2 = 2\cos 36^\circ$.  
A: Noting that a diagonal of a pentagon is $\frac{1+\sqrt5}{2}$, we can draw a diagonal from 2 non-adjacent verticies of a unit pentagon and get a triangle with degree measures $36,36,108$ and side lengths $1,1,\frac{1+\sqrt5}{2}$. Now if we take a cosine of a 36 degree angle, we get that $\cos(36^\circ)=\frac{\frac{1+\sqrt5}{2}}{1}=\frac{1+\sqrt5}{2}$. You can find the other solution by taking $180^\circ-36^\circ$
