Solving $4\cos2x-\sqrt3\cot2x=1$ How would I solve this trigonometric equation?
$$4\cos2x-\sqrt3\cot2x=1$$
I got to this stage, is it a dead end? $$4\sin2x-\sqrt3 = \tan2x$$
 A: $\displaystyle4\cos2x-\sqrt3\cot2x=1\iff2\cos2x\sin2x=\sin\dfrac\pi3\cos2x+\cos\dfrac\pi3\sin2x$
$\displaystyle\implies\sin4x=\sin\left(\dfrac\pi3+2x\right)$
$\displaystyle\implies4x=n\pi+(-1)^n\left(\dfrac\pi3+2x\right)$ where $n$ is any integer
See also : Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $
A: I don't see an easy way out here. Writing
$$
\cos 2x=2\cos^2x-1=\frac{2}{1+\tan^2x}-1
$$
and
$$
\cot 2x=\frac{1}{2}\Bigl(\frac{1}{\tan x}-\tan x\bigr)
$$
we end up with the equation (where $u=\tan x$)
$$
\sqrt{3}u^4-10u^3+6u-\sqrt{3}=0.
$$
One solution is $u=1/\sqrt{3}$ which will give you nice solutions for $x$ in the end. The other three solutions to the polynomial equation are real as well, but not as nice. I write them down for you, and you can consider yourself how to continue,
$$
\begin{aligned}
u_1&=\sqrt{3}-2\sqrt{3}\cos(\pi/9)+2\sin(\pi/9),\\
u_2&=\sqrt{3}+2\sqrt{3}\cos(\pi/9)+2\sin(\pi/9),\\
u_3&=\sqrt{3}-4\sin(\pi/9).
\end{aligned}
$$
A: Hint:
$\cos 2x=\frac{1-tan^2x}{1+tan^2x} $
and, $\cot 2x=\frac{1-tan^2x}{2\tan x}$
now let $1-tan^2x=a$
then 
$1+tan^2x=2-a$  and
$2\tan x = 2\sqrt{1-a}$
now you have a quadratic in a.
