<s>[INST] Solve $\frac{2}{\sin x \cos x}=1+3\tan x$ Solve this trigonometric equation given that $0\leq x\leq180$
$\frac{2}{\sin x \cos x}=1+3\tan x$
My attempt, 
I've tried by changing to $\frac{4}{\sin 2x}=1+3\tan x$, but it gets complicated and I'm stuck. Hope someone can help me out.
 [/INST] Multiply by LHS and RHS by $\sin x\cos x$ :
$$3\sin ^2x +\sin x\cos x -2 =0 \implies 3\cos^2 x-\sin x\cos x -1 =0$$
$$\implies \frac32 (2\cos ^2 x -1) -\frac{2\sin x \cos x }{2}=\frac{-1}{2} \implies 3\cos {2x} - \sin{2x}=-1$$
Now, squaring both the sides, you get :
$$5 \sin^2{2x}-\sin{2x}-4=0 \implies \sin {2x} =1~ \text{or}~ -4/5$$
Now you may proceed !
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