Consider the system of coupled oscillators $$ \frac{d\theta_1}{dt}=2\pi\epsilon\sin(\theta_2-\theta_1)$$ $$ \frac{d\theta_2}{dt}=2\pi\epsilon\sin(\theta_1-\theta_2)$$
Suppose that at $t=0, \theta_1(0)=\frac{\pi}{2}$ and $\theta_2(0)=0$. Sketch the solutions of the solution of $\theta_1$ and $\theta_2$ on the time interval $0\leq t \leq 10$ for $\epsilon=0, \epsilon >0$ but small, and $\epsilon=1$.
$$\begin{cases} \frac{d\theta_1}{dt}=2\pi\epsilon\sin(\theta_2-\theta_1)\\ \frac{d\theta_2}{dt}=2\pi\epsilon\sin(\theta_1-\theta_2) \end{cases} \qquad\to\qquad \frac{d\theta_1}{dt}+\frac{d\theta_2}{dt}=0 $$ $$\theta_1+\theta_2=C$$ $\theta_1(0)=\frac{\pi}{2}$ and $\theta_2(0)=0$ implies $C=\frac{\pi}{2}$ $$\theta_1+\theta_2=\frac{\pi}{2}\quad\;\quad \theta_2=\frac{\pi}{2}-\theta_1 $$ $$\frac{d\theta_1}{dt}=2\pi\epsilon\sin(\frac{\pi}{2}-2\theta_1)$$ $$\frac{d\theta_1}{dt}=2\pi\epsilon\cos(2\theta_1)$$ I suppose that you can take it from here.