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Solve the following equation on the interval $[0, 2\pi)$:

$$4\sin^2{\theta}-2\sin{\theta}=0$$

$0, \pi, \cfrac{\pi}{6}, \cfrac{5\pi}{6}$

$\cfrac{\pi}{2}, \cfrac{3\pi}{2}, \cfrac{\pi}{6}, \cfrac{5\pi}{6}$

$0, \pi, \cfrac{\pi}{3}, \cfrac{5\pi}{3}$

$0, \pi, \cfrac{\pi}{6}, \cfrac{11\pi}{6}$

$\cfrac{\pi}{2}, \cfrac{3\pi}{2}, \cfrac{2\pi}{3}, \cfrac{4\pi}{3}$