A trigonometry equation: $3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$ 
$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$

What are the steps to solve this equation for $ \theta $?
Because, I  am always unable to deal with the product $\sin \theta \cos \theta$.

$3\sin^2{\theta}-2\cos^2{\theta}+5\sin{\theta}\cos{\theta}=0$
Transfer the $\sin{\theta}\cos{\theta}$ term to the RHS.
$3\sin^2{\theta}-2\cos^2{\theta}=-5\sin{\theta}\cos{\theta} $
Then square both sides.
$(3\sin^2{\theta}-2\cos^2{\theta})^2=(-5\sin{\theta}\cos{\theta})^2$
$9\sin^4{\theta}-12\sin^2{\theta}\cos^2{\theta}+4\cos^4{\theta}=25\sin^2{\theta}\cos^2{\theta}$
Then convert all occurrences of $\sin^2{\theta}$ to $1-\cos^2{\theta}$.
$9(1-\cos^2{\theta})^2-12(1-\cos^2{\theta})\cos^2{\theta}+4\cos^4{\theta}=(1-\cos^2{\theta})\cos^2{\theta}$
This will give an equation which can be solved for $\cos^2{\theta}$,
$50\cos^4{\theta}-55\cos^2{\theta}+9=0$
then for $\cos{\theta}$,
then for $\theta$.
