= Solution
For
$$
g_{\theta\theta}=1,\qquad g_{\phi\phi}=\sin^2\theta,
$$
the nonzero <Christoffel symbols> needed here are
$$
\Gamma^\theta_{\phi\phi}=-\sin\theta\cos\theta,
\qquad
\Gamma^\phi_{\theta\phi}
=\Gamma^\phi_{\phi\theta}=\cot\theta.
$$
On a line of constant longitude, $\dot\phi=0$, and the <geodesic equation> reduces to $\ddot\theta=0$ after choosing an <affine parameter>. Thus every meridian is a geodesic.
For a nonconstant line of constant latitude, $\dot\theta=0$ and the $\theta$ equation requires
$$
-\sin\theta\cos\theta\,\dot\phi^2=0.
$$
Away from the coordinate-degenerate poles, this holds only when
$$
\boxed{\theta=\pi/2},
$$
so only the equator is a constant-latitude geodesic.
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