<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] $\cos(4 t) = 8 \cos(t)^4 - 8 \cos(t)^2 + 1$, 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.
$0 \le c \le 9/4$.
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