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?
Best Answer
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 $-2$.