Solving sin⁡3θ=12 on the interval [0,2π]. I don't understand where some solutions came from.

Write out ${\displaystyle 3\theta :\frac{\pi }{6},\frac{5\pi }{6},\frac{13\pi }{6},\frac{17\pi }{6},\frac{25\pi }{6},\frac{29\pi }{6}}$. Now subtract ${\displaystyle \frac{\pi }{6}}$ from ${\displaystyle \frac{13\pi }{6}}$, and you get ${\displaystyle \frac{12\pi }{6}=2\pi }$. The sine function is periodic, with this period. You get all the other by adding ${\displaystyle 2\pi \text{ or }4\pi }$ to the first solutions.
Maybe a more obvious way of thinking about the problem is to say ${\displaystyle \alpha =3\theta }$, and find all solutions of ${\displaystyle \sin \alpha =\frac{1}{2}}$ in the interval ${\displaystyle [0,6\pi ]}$