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.
 A: Note that $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{1}{\sin x \cos x}$ (this is something I've filed away from seeing it in a lot of "verify this trig identity" type problems) .
So your equation can be written $2(\tan x + \cot x) = 1 + 3 \tan x$, or 
$$2\left(\tan x + \frac{1}{\tan x}\right) = 1 + 3 \tan x.$$
Now that everything is in terms of tangent, you should be able to get it into the form of an equation that's quadratic in $\tan x$ (i.e., has multiples of $\tan^2 x,\, \tan x,$ and a number) and solve; the values of $\tan x$ are nice integers, although you will need inverse trig to get one of the $x$-values.
A: 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.
A: 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!
A: 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 !
A: 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 $u$.
