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.

Multiply both sides by $sin(x)cos(x)$ and rearrange to get $2-3sin^2(x)=sin(x)cos(x)$. 
Square both sides: $4-12sin^2(x)+9sin^4(x)=sin^2(x)cos^2(x)$. 
Substitute $1-sin^2(x)$ for $cos^2(x)$ to get $$
4-12sin^2(x)+9sin^4(x) = sin^2(x) - sin^4(x)$$
$$10sin^4(x)-13sin^2(x)+4=0$$ Substitute $u=sin^2(x)$ to get a quadratic equation, with solution $u=\frac{13\pm\sqrt{169-160}}{20}=\frac{4}{5},\frac{1}{2}$
Then solve for $x$ given $u$.
