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 $\sin \theta \cos \theta$.
 A: Try dividing by $\cos^2\theta$ all the equation...
A: Use the linearisation formulae (inverse of the duplication formulae):
\begin{align*}
&=3\frac{1-\cos2\theta}2+\frac52\sin2\theta-2\frac{1+\cos2\theta}2\\
&=\frac12-\frac52(\cos2\theta-\sin2\theta)=\frac12-\frac{5\sqrt 2}2\cos\Bigl(2\theta-\frac\pi4\Bigr),
\end{align*}
whence
\begin{align*}
3\sin^2\theta+&5\sin\theta\cos\theta-2\cos^2\theta=0\iff\cos\Bigl(2\theta-\frac\pi4\Bigr)=\frac1{5\sqrt2}\\
&\iff 2\theta-\frac\pi4\equiv\pm\arccos\frac1{5\sqrt2}\mod2\pi\\
&\iff \theta\equiv\frac\pi8\pm\frac12\arccos\frac1{5\sqrt2}\mod\pi.
\end{align*}
A shorter method:
First observe we cannot have $\cos\theta=0$, for it would imply $\sin\theta=0$ and we cannot have both. So we can divide the equation by $ \cos\theta$, obtaining:
$$3\tan^2\theta+5\tan\theta-2=0$$
Now the quadratic equation $t^2+5t-2=0$ has discriminant equal to $49$ and  roots $\;\Bigl\{-2,\dfrac13\Bigr\}$. Thus we have to solve
$$\begin{cases}\tan\theta=-2\\\tan\theta=\dfrac13\end{cases}\iff\begin{cases}\theta\equiv-\arctan2\mod\pi\\\theta\equiv\arctan\dfrac13\mod\pi=\dfrac\pi2-\arctan 3\mod\pi\end{cases}$$
A: $3\sin^2{\theta}-2\cos^2{\theta}+5\sin{\theta}\cos{\theta}=0$
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 $\theta$.
A: Factorise this into the quadratic:
$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$
$$(3 \sin \theta - \cos \theta )(sin \theta + 2\cos \theta) = 0$$
This gives 2 equations.
$$3 \sin \theta - \cos \theta = 0$$
$$\sin \theta + 2\cos \theta = 0$$
You should be able to solve it by dividing both by cos. And then solve for tan.
(Make sure you check the cases if $$\cos \theta = 0$$ separately afterwards to see if you missed any solutions when dividing by cos)
A: 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
$$\theta  = {1 \over 2}\left( {\left\{ \matrix{
  \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr 
  \pi  - \arcsin \left( {{1 \over {5\sqrt 2 }}} \right) \hfill \cr}  \right. + 2n\pi  - {{3\pi } \over 4}} \right)$$
A: Following excellent advice by @raul, you write immediately
$$\tan(\theta)=\frac{-5\pm\sqrt{5^2+4\cdot3\cdot2}}{2\cdot3}=\frac13,-2.$$
