<s>[INST] Solving $\sin 3\theta=1/2$ on the interval $[0,2\pi]$. I don't understand where some solutions came from. I am learning precalculus, and I understand how to obtain first two solutions, but I don't understand where did last four solutions came from:

All values of $\theta$ in the interval $[0,2\pi]$ that satisfy $\sin 3\theta=1/2$ are
  $$\theta = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}, \frac{25\pi}{18}, \frac{29\pi}{18}$$

 [/INST] $$\theta \in \left[0, 2\pi\right]\implies 3\theta \in \left[0, 6\pi\right]$$
$$\sin 3\theta = \dfrac{1}{2}$$
$$3\theta = \frac{\pi}{6}+2n\pi, \dfrac{5\pi}{6}+2n\pi, n \in \Bbb Z$$
But $3\theta \in \left[0, 6\pi\right]$, so $$3\theta = \dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{13\pi}{6},\dfrac{17\pi}{6},\dfrac{25\pi}{6},\dfrac{29\pi}{6}$$
Therefore
$$\theta = \dfrac{\pi}{18}, \dfrac{5\pi}{18}, \dfrac{13\pi}{18},\dfrac{17\pi}{18},\dfrac{25\pi}{18},\dfrac{29\pi}{18}$$
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