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 $\sin \theta \cos \theta$.

Here is a general strategy dealing with these kind of problems. Using summation formulas for sin and cos one can easily prove the following identities
$$\left\{ \matrix{
  \cos 2\theta  = {\cos ^2}\theta  - {\sin ^2}\theta  = 2{\cos ^2}\theta  - 1 = 1 - 2{\sin ^2}\theta  \hfill \cr 
  \sin 2\theta  = 2\sin \theta \cos \theta  \hfill \cr}  \right.$$
using these you can write
$$\eqalign{
  & {\cos ^2}\theta  = {{1 + \cos 2\theta } \over 2}  \cr 
  & {\sin ^2}\theta  = {{1 - \cos 2\theta } \over 2}  \cr 
  & \sin \theta \cos \theta  = {{\sin 2\theta } \over 2} \cr} $$
put these into your equation to get
$$3\left( {{{1 - \cos 2\theta } \over 2}} \right) + 5\left( {{{\sin 2\theta } \over 2}} \right) - 2\left( {{{1 + \cos 2\theta } \over 2}} \right) = 0$$
simplify to get
$$\eqalign{
  & {1 \over 2} - {5 \over 2}\cos 2\theta  + {5 \over 2}\sin 2\theta  = 0  \cr 
  & \sin 2\theta  - \cos 2\theta  + {1 \over 5} = 0 \cr} $$
the next step is to combine linear combinations of $\sin 2\theta $ and $\cos 2\theta $. You can do this again using summation formulas
$$\left\{ \matrix{
  \sin x + w\cos x = \sin x + \tan \alpha \cos x = {1 \over {\cos \alpha }}\left( {\sin x\cos \alpha  + \sin \alpha \cos x} \right) = {1 \over {\cos \alpha }}\sin \left( {x + \alpha } \right) \hfill \cr 
  \tan \alpha  = w \hfill \cr}  \right.$$
In this case 
$$\tan \alpha  =  - 1\,\,\, \to \,\,\,\alpha  = {{3\pi } \over 4}\,\,\,\, \to \,\,\,\,\cos \alpha  =  - {1 \over {\sqrt 2 }}$$
and hence you can write
$$ - \sqrt 2 \sin (2\theta  + {{3\pi } \over 4}) + {1 \over 5} = 0$$
or
$$\sin (2\theta  + {{3\pi } \over 4}) = {1 \over {5\sqrt 2 }}$$
and hence
$$2\theta  + {{3\pi } \over 4} = \left\{ \matrix{
  \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr 
  \pi  - \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr}  \right. + 2n\pi \,\,\,\,\,\,\,\,\,\,\,\,\,n = 1,2,3,...$$
which finally gives
$$\theta  = {1 \over 2}\left( {\left\{ \matrix{
  \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr 
  \pi  - \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr}  \right. + 2n\pi  - {{3\pi } \over 4}} \right)$$
