Originally Posted by
HallsofIvy First, you need to know some trig identities- like sin^2x+ cos^2x= 1. In this particular problem, that lets you replace "$\displaystyle sin^2\theta$" with [tex]1- cos^2(\theta)[/itex] so the problem becomes [tex]2cos(\theta)- 4(1- cos^2(\theta))= 4cos^2(\theta)+ 2 cos(\theta)- 4= 0[/itex].
Do you recognize that as a quadratic equation? If you replace "$\displaystyle cos(\theta)$" with "y" you have [tex]4y^2+ 2y- 4= 0. You can use the quadratic formula to solve that, then use a calculator to solve $\displaystyle cos(\theta)= y$ for one value of $\displaystyle \theta$. I don't know if you have learned the graph of cosine or use the "unit circle" but you should know that there are two values of $\displaystyle \theta$, between 0 and 360, such that $\displaystyle cos(\theta)= y$. (That is, if y is between -1 and 1. If not, there is no solution.) There are either 0, or 2, or 4 solutions to this equation between 0 and 360.