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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?

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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 $-2$.

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  • $\begingroup$ if $\tan x=-2$ why can't $x$ be $\pi +\arctan(- 2)$ $\endgroup$ – Umesh shankar May 25 '17 at 10:29
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    $\begingroup$ I think you are right. I made a mistake. Only $-1$ should be rejected. $\endgroup$ – CY Aries May 25 '17 at 10:57
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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 $$\tan x(\tan^3x-3\tan x+2)=0$$

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