# Solving or approximating an equation with radicals and arctan function

I have solved a differential equation recently, which left me with this whopper of inverse function to figure out. I know what $c$ is, I just haven't calculated its exact value based on the initial conditions yet.

(1): $-\sqrt{y} \sqrt{5615673-y}+5615673\cdot\tan^{-1}{\left(\frac{\sqrt{y}}{\sqrt{5615673-y}}\right)}=11133.58 t+c$

This equation looked pretty hopeless at first, but then I saw the connection between this and forms like $\sin x = u$, $\cos x = \sqrt{1-u^2}$

Using that connection I looked for a transformation until I found:

(2): $\sqrt{y} = \sqrt{5615673}\sin\alpha$

Which makes:

(3): $\sqrt{5615673 - y} = \sqrt{5615673}\cos\alpha$

Making those substitutions the equation becomes:

(4): $-5615673\cos\alpha\sin\alpha+56165673\alpha = 11133.58 t + c$

From there, it was simple to get to:

(5): $\sin\beta - \beta = s$

Obviously there's no inverse for $\sin x - x$, but is there another approach to solving the 1st equation?

If not, is there a good technique for approximating this sort of thing, or should I just go back to the differential equation and approximate that?

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$\sin (x)-x$ is strictly monotone, so it certainly does have an inverse. It's probably not elementary, though. – tomasz Aug 13 '12 at 19:55
@tomasz is right here. You might want o look into the related Kepler equation, of which much has been written about. – J. M. Aug 14 '12 at 5:04
I meant that it didn't have an inverse in terms of elementary functions. Sorry about that. – Tony Peterson Aug 14 '12 at 11:09