$If r \sin \theta =3 and r=4(1+ \sin \theta ) where 0<\theta \le 2\pi then\theta =$

$$\begin{gathered} \mathrm{r} \mathrm{sin\theta }=3, \mathrm{r}=4(1+\mathrm{sin\theta }) \\ \mathrm{r} \mathrm{sin\theta }=4\mathrm{sin\theta }(1+\mathrm{sin\theta }) \\ 3=4{\sin }^2\mathrm{\theta }+4\mathrm{sin\theta } \\ 4{\sin }^2\mathrm{\theta }+4\mathrm{sin\theta }-3=0 \\ (2\mathrm{sin\theta }+3) (2\mathrm{sin\theta }-1)=0 \\ 2\mathrm{sin\theta }+3\ne 0 \\ \mathrm{sin\theta }=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \mathrm{\theta }=\frac{\mathrm{\pi }}{6}, \frac{5\mathrm{\pi }}{6} \end{gathered}$$