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.

Move everything to the left hand side and substitute $u=-3\tan x$ to get
\begin{align*}0&=\frac{2}{\sin x\cos x}-1-3\tan x\\ &=2\left(\tan x+\frac 1{\tan x}\right)-1-3\tan x\\ &=-1-\frac{6}{-3\tan x}-\frac{3\tan x}{3}\\ &=\frac u3 -1-\frac 6u \\ &= \frac{(u-6)(u+3)}{3u}\end{align*}
Thus we get that $u=-3$ or $u=6$. The first gives $x=\pi n+\frac \pi 4$ for all $n$, and the second gives $x=\pi n - \arctan 2$ for all $n$.
The solutions that satisfy your conditions are for $n=0$ in the first case and for $n=1$ in the second case.
