Breaking a variable out of a trigonometry equation $A = 2 \pi r^2 - r^2  (2 \arccos(d/2r) - \sin(2 \arccos(d/2r)))$
Given $A$ and $r$ I would like to solve for $d$. However, I get stuck breaking the $d/2r$ out the trig functions.
For context this is the area of two overlapping circles minus the overlapping region. Given a radius and desired area I'd like to be able to determine how far apart they should be. I know $A$ should be bounded below by $0 (d = 0)$ and above by $2 \pi r^2 (d \le 2r)$.

This is a transcendental equation for $d$ which can't be solved for $d$ in closed form. You can get rid of the trigonometric functions in the last term using
$$\sin2x=2\sin x\cos x$$
and
$$\sin x=\sqrt{1-\cos^2 x}$$
and thus
$$\sin\left(2\arccos\frac d{2r}\right)=\frac dr\sqrt{1-\left(\frac d{2r}\right)^2}\;,$$
but that still leaves $d$ both in the argument of the arccosine and outside.
