Find $\tan x$ if $x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right)$ Find $\tan x$   if $$x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right) \tag{1}$$
First i converted $$\frac{3 \sin 2x}{5+4 \cos 2x}=\frac{6 \tan x}{9+\tan^2 x}$$
So
$$\arcsin\left( \frac{6 \tan x}{9+\tan^2x}\right)=\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right)$$ Now using above result $(1)$ can be written as
$$2x=2 \arctan(2 \tan^2 x)-\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right) \tag{2}$$
But $$2\arctan (\theta)=\arctan \left(\frac{2 \theta}{1-\theta^2}\right)$$
So $(2)$ becomes
$$2x=\arctan \left(\frac{4 \tan^2x}{1-4 \tan^4x}\right)-\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right)$$
Now using $$\arctan(a)-\arctan(b)=\arctan\left(\frac{a-b}{1+ab}\right)$$ i am getting a sixth degree polynomial in $\tan x$.
is there any better approach?
 A: Let $\alpha=\arctan(2\tan^2x)$ and $\beta=\arcsin(\frac{3\sin2x}{5+4\cos2x})$.
\begin{align}
\sin\beta&=\frac{3\sin2x}{5+4\cos2x}\\
\frac{2\tan\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}&=\frac{3(\frac{2\tan x}{1+\tan^2x})}{5+4(\frac{1-\tan^2x}{1+\tan^2x})}\\
\frac{\tan\frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}&=\frac{3\tan x}{9+\tan^2x}\\
3\tan x\tan^2\frac{\beta}{2}-(9+\tan^2x)\tan\frac{\beta}{2}+3\tan x&=0\\
\left(3\tan \frac{\beta}{2}-\tan x\right)\left(\tan \frac{\beta}{2}\tan x-3\right)&=0\\
\tan\frac{\beta}{2}&=\frac{1}{3}\tan x \quad\textrm{or}\quad \frac{3}{\tan x}
\end{align}
Note that $x=\alpha-\frac{1}{2}\beta$.
\begin{align}
\tan x&=\tan\left(\alpha-\frac{1}{2}\beta\right)\\
&=\frac{\tan \alpha-\tan\frac{\beta}{2}}{1+\tan\alpha\tan\frac{\beta}{2}}\\
&=\frac{2\tan^2 x-\frac{1}{3}\tan x}{1+2\tan^2 x(\frac{1}{3}\tan x)} \quad\textrm{or}\quad \frac{2\tan^2 x-\frac{3}{\tan x}}{1+2\tan^2 x(\frac{3}{\tan x})} \\
&=\frac{\tan x(6\tan x-1)}{3+2\tan^3 x} \quad\textrm{or}\quad \frac{2\tan^3 x-3}{\tan x(1+6\tan x)} \\
\end{align}
So we have $\tan x=0$, $\tan^3x-3\tan x+2=0$ or $4\tan^3x+\tan^2x+3=0$.
Solving, we have $\tan x=0$, $1$, $-1$ or $-2$.
Note that $-1$ should be rejected.
$\tan x=-1$ is corresponding to $\tan\frac{\beta}{2}=\frac{3}{\tan x}$. So $\tan\frac{\beta}{2}=-3$, which is impossible as $\beta\in[\frac{-\pi}{2},\frac{\pi}{2}]$.
The answers are $0$, $1$ and $-2$.
A: Where you have left off, 
$$\dfrac{6\tan x}{9-\tan^2x}=\dfrac{2\cdot\dfrac{\tan x}3}{1-\left(\dfrac{\tan x}3\right)^2}$$
Using my answer here, Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$2\arctan\dfrac{\tan x}3=\begin{cases} \arctan\dfrac{6\tan x}{9-\tan^2x} &\mbox{if } \left(\dfrac{\tan x}3\right)^2\le1\iff-3\le\tan x\le3\\ \pi+\arctan\dfrac{6\tan x}{9-\tan^2x} & \mbox{if } \tan x>3\text{ or }\tan x<-3\end{cases}$$
Now the tougher case $\tan x>3$ or $\tan x<-3;$
$$x=\arctan(2\tan^2x)+\dfrac\pi2-\arctan\dfrac{\tan x}3$$
$$\tan x=\tan\left(\dfrac\pi2+\arctan(2\tan^2x)-\arctan\dfrac{\tan x}3\right)$$ $$=-\cot\left(\arctan(2\tan^2x)-\arctan\dfrac{\tan x}3\right)$$
$$\iff\tan x=-\dfrac{1+2\tan^2x\cdot\dfrac{\tan x}3}{2\tan^2x-\dfrac{\tan x}3}$$
On simplification, we should get a cubic equation in $\tan x$
Can you handle the simpler case  $-3\le\tan x\le3$ to reach at $$\tan x(\tan^3x-3\tan x+2)=0$$
