 Solution

For $r=0$, the equilibrium points satisfy $\sin x=0$, so $x_k=k\pi$ for every integer $k$. The derivative of the vector field is $f^{\prime }(x_k)=-\cos (k\pi )=-(-1{)}^k$. ThusBetween an odd multiple and the next even multiple the sine is negative, so the flow points right; between that even multiple and the next odd multiple it points left. A compact phase-line sketch isEvery initial condition strictly between two adjacent repelling points approaches the even multiple between them. The repelling equilibria themselves remain fixed if initialized exactly there.