= Solution
For nonvertical <uniform precession of a heavy symmetric top>, write $\dot\phi=\Omega$ and fix $\theta$. The $\theta$ <Euler-Lagrange equation> reduces to
$$
\boxed{I_1\cos\theta\,\Omega^2-I_3\omega_3\Omega+Mgl=0.}
$$
For $\cos\theta\ne0$ its solutions are
$$
\Omega=\frac{I_3\omega_3\pm\sqrt{I_3^2\omega_3^2-4I_1Mgl\cos\theta}}{2I_1\cos\theta}.
$$
For $0<\theta<\pi/2$, uniform <precession> therefore requires \b[$|\omega_3|\geq2\sqrt{I_1Mgl\cos\theta}/I_3$]. For $\pi/2<\theta<\pi$, the discriminant is positive for every <spin>, including zero. At $\theta=\pi/2$, the condition is $\Omega=Mgl/(I_3\omega_3)$, requiring nonzero <spin>. Once this condition holds, choose $L_z=I_1\Omega\sin^2\theta+L_3\cos\theta$ and $\dot\psi=\omega_3-\Omega\cos\theta$; both cyclic equations are then satisfied. The exactly vertical configurations are also solutions for arbitrary <spin>, but the azimuthal member of the <Euler angles> is degenerate there, so their existence is separate from this nonvertical discriminant condition.
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