= Solution
Time <independence> conserves the <energy> $E=T+mga\cos\theta$. The two cyclic coordinates give
$$
p_\psi=C(\dot\psi+\dot\phi\cos\theta)=Cn,\qquad
p_\phi=A\dot\phi\sin^2\theta+Cn\cos\theta=h.
$$
These are respectively <angular momentum> along the symmetry axis and <angular momentum> along the fixed upward vertical; $n$ is the axial spin rate. For $u=\cos\theta$ and $-1<u<1$, eliminate $\dot\phi$ using $\dot\phi=(h-Cnu)/[A(1-u^2)]$. The <energy> becomes
$$
E=\frac A2\dot\theta^2+\frac C2n^2+mga u+\frac{(h-Cnu)^2}{2A(1-u^2)}.
$$
Multiplying and rearranging gives
$$
\boxed{A^2\dot\theta^2=\frac{A(2E-Cn^2-2mga u)(1-u^2)-(h-Cnu)^2}{1-u^2}}.
$$
To justify the double-root condition, use the <effective potential>
$$
U(u)=\frac C2n^2+mga u+\frac{(h-Cnu)^2}{2A(1-u^2)}.
$$
Eliminating the cyclic coordinates by Routh reduction gives the remaining equation $A\ddot\theta=-dU(\cos\theta)/d\theta$. A constant nonvertical angle therefore has $U'(u_0)=0$ and $E=U(u_0)$. Since $P(u)=2A(E-U(u))(1-u^2)$, both $P(u_0)$ and $P'(u_0)$ vanish. This argument uses the second-order equation; merely differentiating the <energy> identity at a stationary angle would not establish the derivative condition.
Thus \b[steady nonvertical <precession> gives a common root in $(-1,1)$]. The nonvertical qualification is implicit in the denominator used by the source. A top spinning exactly upright or inverted is also possible but corresponds to an endpoint $u=\pm1$, not an interior root; the unrestricted literal claim would need that exception.
The distinction is captured by the <double-root criterion for steady precession of a symmetric top>.
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