## Trigonometry 7th Edition

$2cos(2\theta)-4sin(\theta)=-1$ $2[1-2sin^2(\theta)]-4sin(\theta)=-1$ $2-4sin^2(\theta)-4sin(\theta)+1=0$ $4sin^2(\theta)+4sin(\theta)-3=0$ $sin(\theta)=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}=\frac{-(4) \pm \sqrt{(4)^2 - 4(4)(-3)}}{2(4)}=\frac{1}{2},\frac{-3}{2}$ $sin(\theta)=\frac{-3}{2}=-1.5\;\;\;\;$ $sin(\theta)=\frac{1}{2}$ $\theta=sin^{-1}(\frac{1}{2})$ We know $sin(\theta)$ is positive in quardent $I$ and quardent $II$ $\theta=30^o\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\;\;\theta=180^o-30^o$ $\theta=30^o\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;\;\;\theta=150^o$ So choice $(c)$ is the right solution because $150^o$ is the larger of two solutions.