![\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}](/tpl/images/0495/8996/969fb.png)
As level c2 tell me how to work out
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