<s>[INST] 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?
 [/INST] 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 $180^\circ-36^\circ$
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