Originally Posted by
crosser43 ok, i gotta solve these equations and my answers have to be $\displaystyle 0\leq x \leq 360^o$
$\displaystyle 2sin^2x-3cosx=0$
change $\displaystyle \textcolor{red}{sin^2{x}}$ to $\displaystyle \textcolor{red}{1 - cos^2{x}}$ and solve the resulting quadratic
and
$\displaystyle 2cosx+3tanx=0$
multiply every term by $\displaystyle \textcolor{red}{\cos{x}}$ , then change $\displaystyle \textcolor{red}{\cos^2{x}}$ to $\displaystyle \textcolor{red}{1 - \sin^2{x}}$ ... same drill, solve the resulting quadratic.