The geodesic equation of a particle is described as $$\left(1-\frac{2\mu}{r}\right)\dot{t}=k \tag{1}$$ $$\left(1-\frac{2\mu}{r}\right)^{-1}\ddot{r}+\frac{\mu c^2}{r^2}\dot{t}-\left(1-\frac{2 \mu}{r}\right)^{-2}\frac{\mu}{r^2}\dot{r}^2-r(\dot{\theta}^2+sin^2\theta\dot{\phi}^2)=0 \tag{2}$$
$$\ddot{\theta}+\frac{2}{r}\dot{r}\dot{\theta}-sin\theta cos\theta\dot{\phi}^2=0 \tag{3}$$ $$r^2sin^2\theta\dot{\phi}=h \tag{4}$$ where $r$ is the radial coordinate, $\theta$ denotes the angle to the z axis ,$\phi$ is the polar angle, $k, \mu$ and $h$ are some constants. $\dot{t}$ denotes the derivative of t with respect to some parameter $\sigma$.
The solution to (3) is $\theta=\pi/2$.
So the third term of (3) vanishes but how do the first and second term yield zero?
$\theta=\pi/2$ is definitely one of the solutions (or at least, a part of it). However, this is a differential equation, and differential equations always have an infinite number of solutions based on their initial conditions. In this case, you don't even have enough equations to do that.
You have three unknown functions: $r(\sigma)$, $\theta(\sigma)$ and $\phi(\sigma)$, so you need three differential equations, and also initial values and derivatives for each of the coordinates (6 numbers for initial condition). Only then you can actually get a specific solution.