What is the correct way to divide a circle in 3 equal-area pieces by dividing it with 2 parallel lines? 
What is the correct way to divide a circle in 3 equal-area pieces by dividing it with 2 parallel lines?

I came up with this question and found a way to solve it but want to know if my attempt at a solution is correct and if it is the most efficient solution. Other solutions are also welcome!
Essentially the question is, on a circle centered at the origin, divide it into 3 equal-area pieces by drawing 2 parallel lines through it, lines perpendicular to the x axis. Where should the lines be drawn? 
For simplicity, I use a unit circle to solve this, and first I rearranged the graph into a function: f(x) = sqrt(1-x^2). Since we can only use the positive or negative square root at a time, I decided to use the positive, and the new area of this semicircle is pi/2, and 1/3 of that would be pi/6. So I created an integral from -1 to x and set it equal to pi/6, and x turned out to be somewhere around +/-0.265. 
Is this a valid solution? I would love to see other ways to solve it. Thank you!
 A: Well, you're right. The problem boils down to solving the trascendental equation
$$ 2\int_{x}^{1}\sqrt{1-z^2}\,dz=\arccos(x)-x\sqrt{1-x^2}=\frac{\pi}{3} $$
which on its turn is equivalent to
$$ \theta-\frac{1}{2}\sin(2\theta) = \frac{\pi}{3} $$
or to $\varphi-\sin\varphi = \frac{2\pi}{3}$ (an instance of Kepler's equation). A step of Newton's method with starting point $\varphi_0=\pi$ leads to the approximated solution $\varphi=\frac{5\pi}{6}$, from which $\theta=\frac{5\pi}{12}$ and 
$$x\approx \cos\frac{5\pi}{12}=\frac{1}{2}\sqrt{2-\sqrt{3}}\approx 0.2588. $$
Further steps of Netwon's method lead to the more accurate approximation $0.264932\approx\frac{448}{1691}$.
A: We can construct a coordinate system so that the circle is given by $x^2+ y^2= R^2$.  Without loss of generality we can take the "parallel lines" to be parallel to the y-axis, say at x= a and x= b with $-R\le a\le b\le R$.  Then the three area are 
$A_1= 2\int_{-R}^a \sqrt{R^2- x^2}dx$
$A_2= 2\int_a^b \sqrt{R^2- x^2}dx$ and
$A_3= 2\int_b^R \sqrt{R^2- x^2}dx$.
Do each of those integrals using the substitution $x= sin(\theta))$, set $A_1= A_2$ and $A_2= A_3, and solve for a and  b.
A: By simple trigonometry, the area of a segment of aperture $\theta$ of a unit circle is given by 
$$\frac{\theta-\sin\theta}2$$ and you want this to be $\dfrac{\pi}3$.
With a numerical solver, $\theta\approx2.6053256746009$ and your $x$ is $\cos\dfrac\theta2\approx 0.2649320846028.$

An initial approximation can be obtained from the Taylor approximation to the third order,
$$\frac{\theta^3}{12}\approx\frac\pi3$$
but this is not very accurate because $\theta$ exceeds $1$. We can work with the complementary angle $\phi:=\pi-\theta$, giving after rework
$$\phi-\frac{\phi^3}{12}\approx\frac\pi6$$
then $\phi\approx0.53646$ and $\theta\approx2.60513$ by resolution of the cubic.
A: Starting from Jack D'Aurizio's answer
$$\theta-\frac{1}{2}\sin(2\theta) = \frac{\pi}{3}$$ Let $\theta=\frac x2$ to make
$$x-\sin(x)=\frac{2\pi}3$$ and use the approximation
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ which was proposed by Mahabhaskariya of Bhaskara I, an Indian mathematician, more than $1400$ years ago.
This leads to the cubic equation
$$4 x^3+\left(16-\frac{20 \pi }{3}\right) x^2+\left(\frac{23 \pi ^2}{3}-16 \pi
   \right) x-\frac{10 \pi ^3}{3}=0$$ which has only one real root and then it can be solved using the hyperbolic method. 
The analytical solution is quite ugly but evaluates to $x \approx 2.60528$ while the exact solution is $x \approx 2.60533$.
A faster, almost as accurate, approximation could be obtained building the $[2,2]$ Padé approximant
$$x-\sin(x)-\frac{2\pi}3=\frac {\frac{\pi }{3}+2 (x-\pi )+\frac{\pi}{36}   (x-\pi )^2 } { 1+\frac{1}{12} (x-\pi )^2}$$ and the numerator cancels when
$$x=\frac{-72+2 \pi ^2+\sqrt{5184-48 \pi ^2}}{2 \pi }\approx 2.60545 $$
