Consider a double cone (like a past and future light cone) centered at the origin. Now imagine a sphere centered at the origin.
What angle of the slope of the double cone makes it so that the it splits the sphere into 3 pieces such that the volume(s) inside the double-cone is equal to the volume outside the double cone?
(This is easier in 2 dimensions where the answer is simply 45 degrees!)
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The volume of the portion of a cone inside a unit sphere equals the solid angle it subtends. You want this angle to be $\dfrac\pi3$ steradians.
From Wolfram, the solid angle corresponding to the half aperture $\theta$ is
$$\pi(2(1-\cos\theta)+\sin\theta).$$
You can solve for $\theta$.
Note that by symmetry, for the double cone to split the sphere into two pieces of equal volume, each hemisphere must also be split into two pieces of equal volume by each cone. So we can just work with the upper hemisphere. Since a sphere has volume $\displaystyle\frac{4\pi r^{3}}{3}$, the volume a hemisphere is $\displaystyle\frac{2\pi r^{3}}{3}$.
Using standard spherical coordinates $(\theta,\phi,\rho)$, the equation of the sphere is $\rho=r$ and the equation of the cone is $\phi=\phi_{0}$, where $\phi_{0}\in[0,\pi/2]$ is the desired angle. We can then easily compute the volume of the sphere below the cone: \begin{align} V &= \int_{0}^{2\pi}\int_{0}^{\phi_{0}}\int_{0}^{r}\rho^{2}\sin{\phi}\ d\rho\ d\phi \ d\theta\\ &=\frac{r^{3}}{3}\int_{0}^{2\pi}\left(-\cos{\phi}\bigg\rvert^{\phi_{0}}_{0}\right)\ d\theta \\ &= \frac{2\pi r^{3}}{3}(-\cos{\phi_{0}}+1) \end{align} Now, we want this volume to be half of the volume of the hemisphere, so we should have $$\frac{2\pi r^{3}}{3}(-\cos{\phi_{0}}+1)=\frac{\pi r^{3}}{3}$$ which yields $$\cos{\phi_{0}}=\frac{1}{2}$$ so the desired angle is $$\phi_{0}=\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$$ I should mention that this angle $\phi$ is measured from the top of the $z$-axis.