[tex]\bf [1-tan(\theta)][2sin(\theta)+1]=0\implies
\begin{cases}
1-tan(\theta)=0\\\\
1=tan(\theta)\\\\
tan^{-1}(1)=\theta\\\\
\qquad \frac{\pi }{4},\frac{5\pi }{4}\\
----------\\
2sin(\theta)+1=0\\\\
2sin(\theta)=-1\\\\
sin(\theta)=\frac{-1}{2}\\\\
sin^{-1}\left(-\frac{1}{2} \right)=\theta\\\\
\qquad \frac{7\pi }{6},\frac{11\pi }{6}
\end{cases}[/tex]
