Expressing the four roots of a particular quartic in terms of trigonometric functions I know one root of the equation (eq.1),
$x^4+ax^3+2x^2-ax+1 = 0$
is,
$x_1 = \tan\big(\tfrac{1}{4}\arcsin(\tfrac{4}{a})\big)$
How to find the other three roots of eq.1 expressed similarly in terms of trigonometric and/or inverse trigonometric functions?
 A: I wanted to find a clever solution in terms of the tangent being the slope of some line crossing an algebraic curve... no luck. This is what I got instead: 
$$\begin{split}x^4+2x^2+1&=a(x-x^3)\\
(x^2+1)^2&=a(x-x^3)\end{split}\tag1$$
Now substitution $x=\tan t$ is natural on the left, because $x^2+1=1/\cos^2 t$.
$$\begin{split}\frac{1}{\cos^4 t}&=a(\tan t-\tan^3 t)\\
1 &=a(\sin t\cos^3t-\sin^3 t\cos t) \end{split}\tag2$$
The rest flows easily: factor out $\sin t \cos t$ and turn it into $\frac12 \sin 2t$, then use $\cos^2 t-\sin^2 t=\cos 2t$, finally arriving at 
$$1=\frac{a}{4}\sin 4t \tag3$$
Now it's time to pay attention to domains: the substitution $x=\tan t$ is a bijection between $\mathbb R$ and $(-\pi/2,\pi/2)$. In the interval $(-\pi/2,\pi/2)$, which is two periods of the function $\sin 4t$, equation (3) has 


*

*no roots if $|a|<4$

*two (double) roots if $|a|=4$

*four roots if $|a|>4$


It is awkward to write down the roots keeping them all in  $(-\pi/2,\pi/2)$. Since all we care about is $\tan t$, adding or subtracting a multiple of $\pi$ is acceptable. For example: $\theta$, $\theta+\pi/2$, $\pi/4-\theta$, and $3\pi/4-\theta$, where $\theta=\arcsin (4/a)$.
