Expressing the roots of $x^4+ax^3+2x^2-ax+1 = 0$ in terms of trigonometric functions

Question

I know one root of the equation

$$x^4+ax^3+2x^2-ax+1 = 0 \tag1$$

is,

$$x_1 = \tan\left(\frac{1}{4}\arcsin\frac{4}{a}\right)$$

How to find the other three roots of eq.1 expressed similarly in terms of trigonometric and/or inverse trigonometric functions?

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Context:

This 13-year old question was given a SECOND answer yesterday and then was promptly closed by someone(s) seeking context.

We will provide context. To solve the general quintic using elliptic functions, one has to solve a quartic in the elliptic modulus $k$, which is essentialy the quartic above as described in this old post after minor scaling changes in the variable $a$. And that is the context of this quartic, as it leads to the solution of the general quintic in elliptic functions.

Answer 1

Given the quartic
$$ x^4 + a x^3 + 2x^2 - a x + 1 = 0 $$ we can use a substitution similar to the one used to solve palindromic quartics
$$ z = x - \frac{1}{x} $$
We first divide by $x^2$ (assuming $x \ne 0$) to obtain
$$ x^2 + a x + 2 - \frac{a}{x} + \frac{1}{x^2} = 0 $$
Grouping terms gives
$$ x^2 + \frac{1}{x^2} + a\left(x - \frac{1}{x}\right) + 2 = 0 $$
and using the substitution
$$ x^2 + \frac{1}{x^2} = z^2 + 2 $$
we obtain
$$ z^2 + a z + 4 = 0 $$
therefore reducing the quartic to a quadratic in $z$. The four solutions of the original equation are given by the equations

$$x^2-z_{i}x-1=0$$

where $z_{1}$ and $z_{2}$ are the roots of the quadratic $z$ equation.

Given the solution provided by OP, we can set
$$ x_{1} = \tan\theta, \quad \text{with} \quad \theta = \frac{1}{4} \arcsin\left(\frac{4}{a}\right) $$
The other root from the same quadratic is
$$ x_{2}=-\frac{1}{x_{1}} = -\cot\theta= \tan(\theta + \frac{\pi}{2}) $$
Indeed, we have
$$ z_{1}=x_{1} - \frac{1}{x_{1}} = \tan\theta - \cot\theta = -\frac{2}{\tan(2\theta)} $$ Thus, from the $z$ equation, $$ z_{2}=-2\tan(2\theta) $$ and the $x_3$ and $x_4$ are given by

$$x_{3}=\sec(2\theta)-\tan(2\theta), \quad x_{4}=-\sec(2\theta)-\tan(2\theta)$$

Using $\sec$ and $\tan$ double angle formula and simplifying,

$$x_{3}=\cot(\theta+\frac{\pi}{4})=-\tan(\theta+\frac{3\pi}{4}), \quad x_{4}=-\tan(\theta+\frac{\pi}{4})$$

Finally

$$x_{1}=\tan(\theta), \quad x_{2}=\tan(\theta+\pi/2), \quad x_{3}=-\tan(\theta+\pi/4), \quad x_{4}=-\tan(\theta+3\pi/4)$$

Answer 2

Actually they are all rendered by the formula

$x=\tan[\frac14\arcsin(\frac4{a})]$

using differemt branches of the $\arcsin$ function. If $\theta$ represents the principal branch value conventionally called $\arcsin u$ for some argument $u$, the periodicity and symmetry relations of the sine function imply that the entire set of branches is given by

$n\pi\pm\theta,$

where $n\in\mathbb{Z}$ and the $\pm$ sign is positive for even $n$, negative for odd $n$.

When $n$ is incremented by $4$, your argument $\frac14\arcsin(\frac4{a})$ is incremented by $\pi$, which matches the fundamental period of the tangent function. So the distinct roots are rendered by putting in $n=0,1,2,3$. The princcipal branch corrssponds to $n=0$. Then for $n=1$ render

$x=\tan[\frac{\pi}4-\frac14\arcsin(\frac4{a})],$

which may be expanded with the formula for the tangent of a difference to give

$x=\dfrac{1-\tan[\frac14\arcsin(\frac4{a})]}{1+\tan[\frac14\arcsin(\frac4{a}))]}.$

The case $n=3$ may be solved the same way. For $n=2$ the formula for the tangent of a sum involves the undefined $\tan(\frac{\pi}2)$, so use the symmetry relation

$\tan(\frac{\pi}2+\alpha)=-1/\tan(\alpha)$

to proceed.

Answer 3

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 $\theta=\arcsin (4/a)$.