Answer
a. $\frac{\pi}{12}+k\pi$, $\frac{5\pi}{12}+k\pi$
b. $\frac{\pi}{12}$, $\frac{5\pi}{12}$, $\frac{13\pi}{12}$, $\frac{17\pi}{12}$
Work Step by Step
a. $2\sin2\theta=1$
$\sin2\theta=\frac{1}{2}$
$2\theta=\frac{\pi}{6}+2k\pi$, $\frac{5\pi}{6}+2k\pi$
$\theta=\frac{\pi}{12}+k\pi$, $\frac{5\pi}{12}+k\pi$
b. If $\theta=\frac{\pi}{12}+k\pi$, the only solutions in $[0, 2\pi)$ are $\frac{\pi}{12}$ and $\frac{13\pi}{12}$. If $\theta=\frac{5\pi}{12}+k\pi$, the only solutions in $[0, 2\pi)$ are $\frac{5\pi}{12}$ and $\frac{17\pi}{12}$.
