= Solution
The angles $\psi$ and $\phi$ are cyclic. Their <conjugate momentum> components are $p_\psi=I_3(\dot\psi+\dot\phi\cos\theta)$ and $p_\phi=I_1\dot\phi\sin^2\theta+p_\psi\cos\theta$, so both are conserved; consequently $\omega_3=p_\psi/I_3$ is constant. With $q=I_3\omega_3=p_\phi>0$,
$$
\dot\phi=\frac{q(1-\cos\theta)}{I_1\sin^2\theta}=\frac{q}{2I_1\cos^2(\theta/2)}.
$$
Eliminating the cyclic velocities gives the nutation <effective potential>
$$
V_{\rm eff}(\theta)=\frac{q^2}{2I_1}\tan^2(\theta/2)+gl\cos\theta,
$$
apart from the constant axial <spin> energy. Thus
$$
\boxed{\ddot\theta=\frac{gl}{I_1}\sin\theta\left[1-\frac{q^2}{4I_1gl\cos^4(\theta/2)}\right].}
$$
The endpoint $\theta=\pi$ is excluded for these nonzero conserved momenta: the <effective potential> diverges there. The apparent Euler-angle singularity at $\theta=0$ is interpreted by its smooth small-tilt limit.
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