= Solution
The <Poincaré-Bendixson theorem> says that a nonempty <compact> omega-limit set of a smooth planar flow containing no equilibrium is a <periodic orbit>; use its allowed cylinder version here. For $F>1$, choose $p_-=(F-1)/(2k)$ and $p_+=2(F+1)/k$. At the lower boundary $\dot p>0$ and at the upper boundary $\dot p<0$, so the <compact> annulus $p_-\leq p\leq p_+$ is positively invariant. It has no equilibrium because $F-\sin\theta>0$. The theorem supplies a <periodic orbit>, and $\dot\theta=p>0$ makes it encircle the cylinder.
Every <periodic orbit> must lie in $p>0$: for $p\leq0$, $\dot p>0$, and a crossing of $p=0$ can only be upward. In $p>0$ a <periodic orbit> is a <graph of a function> satisfying
$$
\frac{dp}{d\theta}=\frac{F-\sin\theta}{p}-k.
$$
Two distinct <function graphs> cannot cross by uniqueness of this scalar differential equation. If $p_1>p_2$, their difference obeys
$$
(p_1-p_2)'=-\frac{F-\sin\theta}{p_1p_2}(p_1-p_2).
$$
It contracts by a factor strictly less than $1$ during one circuit, contradicting periodicity of both <function graphs>. The scalar return map is order preserving, so a multi-circuit periodic <graph of a function> also cannot have a period larger than one circuit. Therefore \b[there is exactly one <periodic orbit>, and it encircles the cylinder]. The standard auxiliary results used are existence/uniqueness of smooth differential equations and invariance at inward-pointing boundaries.
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