# Exact solutions to an equation containing arctan

Is there a way to solve this equation for x?

$$\sqrt{x^2+8}-x-4\arctan\left(\frac{x+3}{x^2+x+6}\right) = 0$$

Numerically, I get $x \approx 1.39$, but is there a way I can get an exact solution?

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I've converted your equation to LaTeX. – Zev Chonoles Jul 31 '11 at 21:01

The solutions can't be algebraic numbers, because if $x$ was algebraic the same would be true of $y = (\sqrt{x^2+8} - x)/4$ and $z = (x+3)/(x^2 + x + 6)$. Now your equation says $\tan y = z$, and thus $\cos^2 y = \frac{1}{z^2+1}$, so $\cos y$ would be algebraic. But by Lindemann's theorem $\cos y$ is transcendental for any nonzero algebraic $y$.

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+1 I did not think the theory about the existence of algebraic solutions is so strong. – Listing Aug 1 '11 at 10:54

Thanks to Wims Function Calculator, the solution is $$x \approx 1.39370747885618057089248282982,$$ and is not recognized by the Inverse Symbolic Calculator.

EDIT: Wims Function Calculator further gives $$x \approx 1.3937074788561805708924828298190336450650412142437367971287022786222374715488762515547338431838298094439414865124475411133191132591019104894926923243497788034792713333649381694869738035510413209296893.$$ However, the result from the Inverse Symbolic Calculator indicates that a "closed-form" solution is not likely to exist.

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I don't think this answers the question of the OP. – Listing Jul 31 '11 at 21:26
It answers the question in the negative. – Dave Radcliffe Jul 31 '11 at 22:03
@Dave: Thanks for pointing this out. – Shai Covo Jul 31 '11 at 22:06
It's very unlikely that a closed-form solution to this type of equation exists. On the other hand, the Inverse Symbolic Calculator doesn't really rule out moderately complicated "closed-form" numbers. For example, there is no match to $\arctan(\sin(2)+\cos(3)+\tan(4))$. – Robert Israel Jul 31 '11 at 23:00
Thanks, that's what I suspected. – Michael Jul 31 '11 at 23:06