Picture an $x$-axis running at a right angle to the two vertical lines in your picture.
Draw the ray from the center to the point on the pizza in the upper right were your vertical line on the right intersects the boundary. Let $\theta$ be the angle that that ray makes with the $x$-axis.
Similarly draw a ray from the center to the lower right intersection of a vertical line with the boundary, corresponding to the angle $-\theta$.
Then the length of the intersection of that line with the pizza is $2\sin\theta$ (where the radius of the pizza is $1$).
The fraction of the pizza between those rays is $2\theta/(2\pi)$ of the whole pizza (since $2\theta$ is the angle between those bounding rays and $2\pi$ is the angle encompassing the whole pizza.
The part to the right of that vertical line is what you want to be $1/3$ of the pizza. That part is the whole part between those rays minus the part between those rays that is to the left of that line. The part between those rays to the left of that line is a triangle. The area of a triangle is $\frac 1 2\times\text{base}\times\text{height}$. Call that vertical part of length $2\sin\theta$ the base; then the height is $\cos\theta$. The area of the triangle is therefore $\frac 1 2 (2\sin\theta)(\cos\theta)$.
The area to the right of that vertical line is therefore
\begin{align}
& \left( \frac{2\theta}{2\pi}\times\text{area of the whole pizza} \right) - (\sin\theta\cos\theta) \\[10pt]
= {} & \frac\theta\pi\cdot\pi - \sin\theta\cos\theta \\[10pt]
= {} & \theta-\sin\theta\cos\theta\qquad = \theta - \frac 1 2 \sin(2\theta).
\end{align}
You want to make that equal to $1/3$ of the whole pizza, thus to $(1/3)\pi$. You can do that by Newton's method.
