<s>[INST] 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 $\sin \theta \cos \theta$.
 [/INST] 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)
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