Question

Please help AS LEVEL c2 tell me how to work out

Answer 1

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