Answer
( C )
Work Step by Step
$2cos(2\theta)-4sin(\theta)=-1$
$2[1-2sin^2(\theta)]-4sin(\theta)=-1$
$2-4sin^2(\theta)-4sin(\theta)+1=0$
$4sin^2(\theta)+4sin(\theta)-3=0$
$sin(\theta)=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}=\frac{-(4) \pm \sqrt{(4)^2 - 4(4)(-3)}}{2(4)}=\frac{1}{2},\frac{-3}{2}$
$sin(\theta)=\frac{-3}{2}=-1.5\;\;\;\;$
$sin(\theta)=\frac{1}{2}$
$\theta=sin^{-1}(\frac{1}{2})$
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.
