If $0 \le \theta \le 4 \pi,$ find all values of $\theta$ which satisfy
\[\log_{\frac{1}{2} \sin 2 \theta} \sin \theta = \frac{1}{2}.\]Enter all the solutions, separated by commas.
From the given equation,
\[\left( \frac{1}{2} \sin 2 \theta \right)^{\frac{1}{2}} = \sin \theta.\]Squaring both sides, we get
\[\frac{1}{2} \sin 2 \theta = \sin^2 \theta.\]Then $\sin \theta \cos \theta = \sin^2 \theta,$ so
\[\sin \theta \cos \theta - \sin^2 \theta = \sin \theta (\sin \theta - \cos \theta) = 0.\]Thus, $\sin \theta = 0$ or $\sin \theta = \cos \theta.$

If $\sin \theta = 0,$ then $\frac{1}{2} \sin 2 \theta = 0,$ which is not allowed as the base of a logarithm.

Otherwise, $\sin \theta = \cos \theta.$  Then $\tan \theta = 1.$  The solutions to this equation are $\frac{\pi}{4},$ $\frac{5 \pi}{4},$ $\frac{9 \pi}{4},$ and $\frac{13 \pi}{4}.$  But $\sin \theta$ must be positive, in order to take the logarithm of it, so the only solutions are $\frac{\pi}{4}, \frac{9 \pi}{4}.$