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.

Let $\tan\frac{x}{2}=t$.
Hence, we need to solve
$$\frac{2}{\frac{2t}{1-t^2}\cdot\frac{1-t^2}{1+t^2}}=1+\frac{6t}{1-t^2}$$ or
$$t^4+t^3-4t^2-t+1=0$$ or
$$t^4-t^3-t^2+2t^3-2t^2-2t-t^2-t+1=0$$ or
$$(t^2-t-1)(t^2+2t-1)=0,$$
which gives 


*

*$\tan\frac{x}{2}=\sqrt2-1$ or $\frac{x}{2}=22.5^{\circ}+180^{\circ}k$, where $k=0$ or $x=45^{\circ}$;

*$\tan\frac{x}{2}=-\sqrt2-1$, which is impossible;

*$\tan\frac{x}{2}=\frac{1+\sqrt5}{2}$ or $x=2\arctan\frac{1+\sqrt5}{2}$;

*$\tan\frac{x}{2}=\frac{1-\sqrt5}{2}$, which is impossible.
Id est, the answer is $\left\{\frac{\pi}{4},  2\arctan\frac{1+\sqrt5}{2}\right\}$.
Done!
