$\frac{\arccos\left(1\ - \frac{3}{2r^2}\right)}{2\pi}\cdot\pi r^2=\left(r^2-\frac{3}{4}\right)\left(\frac{3}{4}\right)\ + \frac{\sqrt{3}}{4}-\frac{\pi}{12}$
Background: I'm calculating the radius of a circle with a chord of length $\sqrt{3}$ and the relative segment of area $\frac{\sqrt{3}}{4}-\frac{\pi}{12}$. Here is an image for it.
Notes: This is not a homework question, I'm just doing it to design an icon.
Thanks!
As written, I am afraid that there is solution for $r$.
If $360$ stands for $2\pi$, this is a different story. Assuming it, you then look for the zero of function $$f(r)=\frac{1}{2} r^2 \cos ^{-1}\left(1-\frac{3}{2 r^2}\right)-\left(\frac{3}{4} \left(r^2-\frac{3}{4}\right)+\frac{\sqrt{3}}{4}-\frac{\pi }{12}\right)$$ which will not show any explicit solution.
Graphing, we can see that the solution is quite close to $\frac \pi 2$. So, to get an approximation, use Taylor series to get $$f(r)=\frac{1}{48} \left(27-12 \sqrt{3}+4 \pi -9 \pi ^2+6 \pi ^2 \cos ^{-1}\left(\frac{\pi ^2-6}{\pi ^2}\right)\right)+\frac{1}{4} \left(r-\frac{\pi }{2}\right) \left(-3 \pi -2 \pi \sqrt{\frac{3}{\pi ^2-3}}+2 \pi \cos ^{-1}\left(\frac{\pi ^2-6}{\pi ^2}\right)\right)+O\left(\left(r-\frac{\pi }{2}\right)^2\right)$$ and ignoring the higher order terms this will give (as an approximation)= $$r=\frac \pi 2+\frac{27-12 \sqrt{3}+4 \pi -9 \pi ^2+6 \pi ^2 \cos ^{-1}\left(\frac{\pi ^2-6}{\pi ^2}\right)}{12 \pi \left(3+2 \sqrt{\frac{3}{\pi ^2-3}}-2 \cos ^{-1}\left(\frac{\pi ^2-6}{\pi ^2}\right)\right)}\approx 1.55899$$
If you need more accuracy, just use Newton method to obtain the following iterates $$\left( \begin{array}{cc} n & r_n \\ 0 & 1.57079632679 \\ 1 & 1.55899274762 \\ 2 & 1.55893121417 \\ 3 & 1.55893121250 \end{array} \right)$$
Edit
A more efficient way would be to let $r=\sqrt{\frac{3}{2 (1-x)} }$ which makes the function to be $$g(x)=a + b x - \cos^{-1}(x)$$ where $$a=\frac{3}{4}+\frac{1}{\sqrt{3}}-\frac{\pi }{9}\qquad \text{and} \qquad b=\frac{3}{4}-\frac{1}{\sqrt{3}}+\frac{\pi }{9}$$ Function $g(x)$ really looks like a straight line.
Expanded as a Taylor series $$g(x)=\left(a-\frac{\pi }{2}\right)+(b+1) x+\frac{x^3}{6}+O\left(x^5\right)$$ which has one real root. Solving for it leads to $x=0.38320763$ to which corresponds $r=1.5594668$ which is not bad.
Being lazy, using Newton method with $x_0=0$ would give the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.0000000000 \\ 1 & 0.3893709937 \\ 2 & 0.3827904140 \\ 3 & 0.3827837464 \end{array} \right)$$