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 $\sin \theta \cos \theta$.

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
$$\begin{cases}\tan\theta=-2\\\tan\theta=\dfrac13\end{cases}\iff\begin{cases}\theta\equiv-\arctan2\mod\pi\\\theta\equiv\arctan\dfrac13\mod\pi=\dfrac\pi2-\arctan 3\mod\pi\end{cases}$$
