of  satisfies $x(t) < x_0$ for all $t>t_0$, so that $\cS(\epsilon)< x_0$ for all $\epsilon<\epsilon^-$. By continuity of solutions with respect to parameters, the map $\cS$ is continuous, and therefore there exists $\epsilon \in [\epsilon^-,\epsilon^+]$ such that $\cS(\epsilon) =
x_0$. The corresponding solution $x(t)$ is $T$-periodic. The same comparison argument shows that if $0<\epsilon_1 < \epsilon_2$, and for $i=1,2$, $x_i(t)$ is the solution of  with $\epsilon = \epsilon_i$, then $x_1(t_0+T) < x_2(t_0+T)$, so that $\cS(\epsilon_1) <
\cS(\epsilon_2)$. In other words, $\cS(\epsilon)$ is a monotonically increasing function, and so the value of $\epsilon$ above is unique. This concludes the proof of part (i). Now suppose that $K^+ /
K^- < (\cos x_0)^{-1}$, and let $\epsilon \in [\epsilon^-,\epsilon^+]$ be such that the solution $x(t)$ of  with initial condition $x(t_0)=x_0$ is $T$-periodic, as in part (i). Let $x^+ =
\cos^{-1}\big(\frac{K^-}{K^+} c_{x_0}\big)$, and consider the subset of the extended phase space $S^+= \{(x,t): x > x^+\}$. This subset is positively invariant since $\tilde a>0$ and $$\begin{aligned}
\dot x\big|_{x=x^+} &= -\tilde a(t) \frac{K^-}{K^+} \cos x_0 +
\epsilon \tilde \mu(t) \\
&\hspace*{-5ex}\geq -\tilde a(t)
\frac{K^-}{K^+} \cos x_0 + (\cos x_0 / K^+) \tilde \mu(t)\\
&\hspace*{-5ex}\geq \frac{\tilde a(t)}{K^+} \cos x_0 \left( -K^- +
\tilde \mu(t)/\tilde a(t)\right)\\
&\hspace*{-5ex}\geq 