Just change the fraction into an equivalent in the sine function.
[tex]sin(2\theta)= \frac{1}{2} \\ sin(2\theta)=sin( \frac{ \pi }{6}+2k \pi ) \\ 2\theta=\frac{ \pi }{6}+2k \pi \\ \theta=\frac{ \pi }{12}+k \pi \\ \\ sin(2\theta)= \frac{1}{2} \\ sin(2\theta)=sin( \frac{ 15\pi}{18}+2k \pi ) \\ 2\theta=\frac{ 15\pi }{18} +2k \pi \\ \theta=\frac{ 15\pi }{36}+k \pi [/tex]
Therefore:
[tex]\boxed {S=(\theta \epsilon R | \theta=\frac{ \pi }{12}+k \pi ~or~ \theta=\frac{ 15\pi }{36}+k \pi, k \epsilon Z)}[/tex]
If you notice any mistake in my english, please let me know, because i am not native.
