please help me with this problem! in the box I put my answer, however, I got it wrong so if possible, please help me correct whatever my mistake was!

Given
[tex]\sin (2\theta)=\frac{1}{2}[/tex][tex]\begin{gathered} 2\theta=sin^{-1}(\frac{1}{2})\text{ } \\ 2\theta=\frac{\pi}{6} \\ divide\text{ both side by 2} \\ \frac{2\theta}{2}=\frac{\pi}{6}\div2 \\ \\ \theta=\frac{\pi}{6}\text{ x }\frac{1}{2} \\ \\ \theta=\frac{\pi}{12}\text{ } \end{gathered}[/tex]But sine is also positive in the second quadrant.
Therefore, the solution to the equation also lie in the second quadrant.
Hence,
[tex]\begin{gathered} \theta=\pi-\frac{\pi}{12} \\ \theta=\frac{12\pi-\pi}{12} \\ \theta=\frac{11\pi}{12} \end{gathered}[/tex]The solution of the equation is
[tex]\theta=\frac{\pi}{12}\text{ and }\theta=\frac{11\pi}{12}[/tex]