Suppose I have a semi-circle (or a full circle, as I don’t think it matters) of radius 1 and centred at (0, 0). Ie. I have half of a unit circle cantered on the origin.
How do I find the x-axis values where I divide the semicircle into parts with equal areas using parallel vertical lines?
Taken from the link, tou need to find the zeros of the function $$f(x)=x - \sin(x) - a\qquad \text{where} \qquad a=\frac{2\pi k}{n+1}$$ and only a numerical method would do except if you want some approximation.
Suppose that $x$ is small; expand as Taylor series $$x-\sin(x)=\frac{x^3}{6}-\frac{x^5}{120}+\frac{x^7}{5040}-\frac{x^9}{362880}+O\left(x^{11}\right)$$ Now, using series reversion $$x=t+\frac{t^3}{60}+\frac{t^5}{1400}+\frac{t^7}{25200}+O\left(t^{9}\right)\qquad \text{where} \qquad t=\sqrt[3]{6a}$$ Trying for $n=6$, we have $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 1.85645 & 1.85694 \\ 2 & 2.43566 & 2.44036 \\ 3 & 2.89727 & 2.91624 \\ 4 & 3.31189 & 3.36694 \\ 5 & 3.70517 & 3.84283 \end{array} \right)$$
Not perfect but not so bad (I hope). If you want to polish the root, use Newton method with these estimates. For $n=6$ and $k=5$, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 3.705173789 \\ 1 & 3.839890315 \\ 2 & 3.842826124 \\ 3 & 3.842827697 \end{array} \right)$$
We could do it better taking more terms in the series expansion and trensform the inverse series as a Padé approximant; this would give $$x=t \,\,\frac{ 1-\frac{1493 }{21120}t^2+\frac{167 }{268800}t^4} {1-\frac{123 }{1408}t^2+\frac{403 }{295680}t^4 } \qquad \text{where} \qquad t=\sqrt[3]{6a}$$ For the worked example, the results would be $$\{1.85693,2.44026,2.91548,3.36304,3.82569\}$$
If we start from @SteeveDroz's answer $$\int_{0}^{v_i} \sqrt{-x^2 + 2 x r}\,dx =\frac {\pi r^2 }{2n}i \quad\text{with }\quad 0 < i \leq n$$ the lhs write $$\frac{1}{2} \sqrt{v_i (2 r-v_i)} \left(\frac{2 r^2 \tan ^{-1}\left(\frac{\sqrt{v_i}}{\sqrt{2 r-v_i}}\right)}{ \sqrt{v_i(2 r-v_i)}}-r+v_i\right)$$ Let $v_i=r\,k_i $ to make the equation $$(k_i-1)\sqrt{k_i(2-k_i) } +2 \tan ^{-1}\left(\sqrt{\frac{k_i}{2-k_i}}\right)=\frac {i\pi }{n}$$
So, we are looking for the zero of the very smooth function $$f(x)=(x-1)\sqrt{x(2-x) } +2 \tan ^{-1}\left(\sqrt{\frac{x}{2-x}}\right)-a$$ $$f'(x)=2 \sqrt{x(2-x) }$$ which can easily be solved using Newton method.
Edit
We can approximate the solution using a Taylor expansion around $x=1$ and get for $$(x-1)\sqrt{x(2-x) } +2 \tan ^{-1}\left(\sqrt{\frac{x}{2-x}}\right)$$ $$\frac{\pi }{2}+2 (x-1)-\frac{1}{3} (x-1)^3-\frac{1}{20} (x-1)^5-\frac{1}{56} (x-1)^7-\frac{5}{576} (x-1)^9+O\left((x-1)^{11}\right)$$ Now, using series reversion $$x=1+t+\frac{t^3}{6}+\frac{13 t^5}{120}+\frac{493 t^7}{5040}+\frac{37369 t^9}{362880}+O\left(t^{11}\right) \quad\text{with }\quad t=\frac{1}{2} \left(a-\frac{\pi }{2}\right)$$ which seems to be quite decent $$\left( \begin{array}{ccc} a & \text{approximation} & \text{solution} \\ 0.25 & 0.27017 & 0.26739 \\ 0.50 & 0.43265 & 0.43246 \\ 0.75 & 0.57659 & 0.57659 \\ 1.00 & 0.71051 & 0.71051 \\ 1.25 & 0.83890 & 0.83890 \\ 1.50 & 0.96459 & 0.96459 \\ 1.75 & 1.08972 & 1.08972 \\ 2.00 & 1.21630 & 1.21630 \\ 2.25 & 1.34668 & 1.34668 \\ 2.50 & 1.48422 & 1.48426 \\ 2.75 & 1.63479 & 1.63542 \\ 3.00 & 1.80992 & 1.81865 \end{array} \right)$$
As it can be seen, this is good except for small values of $a$. For such a case, expanding again but around $x=0$, series reversion leads to $$x=t+\frac{t^2}{10}+\frac{11 t^3}{350}+\frac{823 t^4}{63000}+O\left(t^{5}\right)\quad\text{with }\quad t=\frac{1}{2} \left(\frac{3a}{2}\right)^{2/3}$$ $$\left( \begin{array}{ccc} a & \text{approximation} & \text{solution} \\ 0.05 & 0.089736 & 0.089736 \\ 0.10 & 0.143241 & 0.143242 \\ 0.15 & 0.188601 & 0.188603 \\ 0.20 & 0.229477 & 0.229481 \\ 0.25 & 0.267383 & 0.267392 \\ 0.30 & 0.303129 & 0.303145 \end{array} \right)$$