Optimum angle for maximum torque in an electric motor taking back emf into account (once it stabilizes)

The standard answer is that maximum torque is when the coil is horizontal, but this assumes a constant current flowing through the coil. When the coil is horizontal, the back emf is a maximum so there are a minimum current and magnetic force. These two effects are in opposition, so I’m looking for a more rigorous proof.

Let $theta$ be the angle between the horizontal and the coil

$$IR=V_{supplied}-V_{back}=V-frac{d(nBAsin{theta)}}{dt}=V-nBAcos{theta}frac {dtheta}{dt}$$

$$F=nBIl=(frac{nBl}{R})(V-nBAcos{theta}frac{dtheta}{dt})$$

For the torque $T$:

$$T=2Frcos{theta}=(frac{2nBl}{R})cos{theta}(V-nBAcos{frac{dtheta}{dt}})=Ifrac{d^2theta}{d^2t}$$

$$T_{max}rightarrow T’=0rightarrow-Vsin{theta}+2nBAcos{theta}sin{frac{dtheta}{dt}}-nBA{cos}^2thetafrac{dtheta}{dt}=0rightarrow$$

$$sin{theta}(2nBAcos{theta}frac{dtheta}{dt}-nBAfrac{d^2theta}{d^2 t}+nBAsin{theta}frac{d^2theta}{d^2 t}-V)=0$$

$sin{theta}=0 rightarrow theta=0$, $theta=pi$ OR $2nBAcos{theta}frac{dtheta}{dt}-nBAθfrac{d^2theta}{d^2 t}+nBAsinfrac{d^2theta}{d^2 t}-V=0$

This is the standard answer; maximum torque when the coil is horizontal. How would I verify that this is indeed a maximum stationary point and how would I solve the second possible solution for the stationary points?