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I know one root of the equation (eq.1),

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


$x_1 = \tan\big(\tfrac{1}{4}\arcsin(\tfrac{4}{a})\big)$

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

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There are three non-principal values of the arcsine that have different tangents when divided by 4 -- namely $\theta+2\pi$, $\pi-\theta$ and $3\pi-\theta$, where $\theta$ is the principal arcsine. If $\tan(\theta/4)$ is a root, chances are excellent that the others will be, too. – Henning Makholm Feb 11 '12 at 18:00
Thanks for the tip. I found that, given $x = \tan(\theta/4)$, then two roots are given by $\theta = \pi/2 +\arcsin\big(\pm\sqrt{1-(4/a)^2}\big)$, and the other two are $\theta = -3\pi/2 +\arcsin\big(\pm\sqrt{1-(4/a)^2}\big)$. (I forgot to mention that a is non-zero.) – Tito Piezas III Feb 11 '12 at 19:06
up vote 1 down vote accepted

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

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