inverse functions when solving trig equations. I am in Pre-calc this semester, and I have been given the problem:
$\cos(\theta) = -\frac12$
The answers I get are $30^o$ and $300^o$ but the answers I get from the homework key are $120^o $ and $240^o$. Why do the inverse equations give me answers that arent correct? Is there a better way to solve these problems? My work is as follows:
$\cos(\theta)=x$
$x= -\frac12$
$x^2+y^2=1$
$(-\frac12)^2+y^2=1$
$\frac14+y^2=1$
$y^2=\frac34$
$\sqrt y^2=\pm\sqrt\frac34$
$y=\pm\frac{\sqrt3}{2}$
$\sin(\theta)=y$
$\sin(\theta)=\frac{\sqrt3}{2}$
$\sin(\theta)=-\frac{\sqrt3}{2}$
$\sin^{-1}(\frac{\sqrt3}{2})=\theta$
$\sin^{-1}(-\frac{\sqrt3}{2})=\theta$
$\theta=60^o,300^o$

Since $\cos\theta=-\frac{1}{2}$ and $\sin\theta=\pm\frac{\sqrt 3}{2}$ implies that $\theta$ lies in the quadrant second or third, so $\theta$ is $\pi\mp\frac{\pi}{3}=\frac{2\pi}{3}$  or $\frac{4\pi}{3}.$
