Solve the equation $\cos^{-1}\frac{x^2-1}{x^2+1}+\tan^{-1}\frac{2x}{x^2-1}=\frac{2\pi}{3}$ $\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$
Let's first find the domain
$$-1<=\dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1}{x^2+1}>=-1 \text { and } \dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1+x^2+1}{x^2+1}>=0 \text { and } \dfrac{x^2-1-x^2-1}{x^2+1}<=0$$
$$\dfrac{2x^2}{x^2+1}>=0 \text { and } \dfrac{-2}{1+x^2}<=0$$
$$x\in R$$
$$x^2-1\ne0$$
$$x\ne\pm1$$
$$\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$$
$$\pi-\cos^{-1}\dfrac{1-x^2}{1+x^2}-\tan^{-1}\dfrac{2x}{1-x^2}=\dfrac{2\pi}{3}$$
$$\dfrac{\pi}{3}=\cos^{-1}\dfrac{1-x^2}{1+x^2}+\tan^{-1}\dfrac{2x}{1-x^2}$$
Substituting $x$ by $\tan\theta$
$$x=\tan\theta$$
$$\tan^{-1}x=\theta$$
$$\theta\in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)-\{-\dfrac{\pi}{4},\dfrac{\pi}{4}\}$$
$$\dfrac{\pi}{3}=\cos^{-1}\dfrac{1-\tan^2\theta}{1+\tan^2\theta}+\tan^{-1}\dfrac{2\tan\theta}{1-\tan^2\theta}$$
$$\dfrac{\pi}{3}=\cos^{-1}(\cos2\theta)+\tan^{-1}(\tan2\theta)$$
$$2\theta\in(-\pi,\pi)-\{-\dfrac{\pi}{2},\dfrac{\pi}{2}\}$$
Now we break the range of $2\theta$ into various parts:-
Case $1$: $2\theta\in\left(-\pi,-\dfrac{\pi}{2}\right)$,$\theta\in\left(-\dfrac{\pi}{2},-\dfrac{\pi}{4}\right)$
$$\dfrac{\pi}{3}=2\pi+2\theta+\pi+2\theta$$
$$-\dfrac{8\pi}{3}=4\theta$$
$$-\dfrac{2\pi}{3}=\theta$$
But it is not the range of $\theta$ we assumed
Case $2$: $2\theta\in\left(-\dfrac{\pi}{2},0\right]$,$\theta\in\left(-\dfrac{\pi}{4},0\right]$
$$\dfrac{\pi}{3}=-2\theta+2\theta$$
$$\dfrac{\pi}{3}=0 \text { not possible }$$
Case $3$: $2\theta\in\left(0,\dfrac{\pi}{2}\right)$,$\theta\in\left(0,\dfrac{\pi}{4}\right)$
$$\dfrac{\pi}{3}=2\theta+2\theta$$
$$\dfrac{\pi}{12}=\theta$$
It is coming in the range of $\theta$, so its a valid solution.
$$\tan^{-1}x=\dfrac{\pi}{12}$$
$$x=\tan\dfrac{\pi}{12}$$
$$x=2-\sqrt{3}$$
Case $4$: $2\theta\in\left(\dfrac{\pi}{2},\pi\right)$, $\theta\in\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)$
$$\dfrac{\pi}{3}=2\pi-2\theta+2\theta-\pi$$
$$\dfrac{\pi}{3}=\pi \text { not possible }$$
So only solution is $2-\sqrt{3}$, but actual answer is $2-\sqrt{3}, \sqrt{3}$
 A: Let me propose an alternative pathway.
Suppose there exists $y$ such that
$$
\frac{2x}{x^2-1} = \tan(y).
$$
This seems resonable as $\tan$ range is the whole $\mathbb{R}$.
Now observe
$$
\frac{1}{\cos^2(y)} = \frac{\sin^2(y) + \cos^2(y)}{\cos^2(y)} = 1+\tan^2(y)
= 1 + \frac{4x^2}{(x^2-1)^2} = \frac{x^4 - 2x^2 + 1+4x^2}{(x^2-1)^2}
= \left(\frac{x^2+1}{x^2-1}\right)^2.
$$
Thus we can choose $y$ such that
$$
\cos(y) = \frac{x^2-1}{x^2+1}.
$$
Now the first equation simplifies to
$$
\cos^{-1}(\cos(y)) + \tan^{-1}(\tan(y)) = \frac{2\pi}{3}.
$$
This is the sum of two periodic functions, one with period $2\pi$ and one with period $\pi$.
Thus it suffices to solve the problem for $[0;2\pi]$ interval to get the full solution.
The problem is $\arccos:[-1;1]\rightarrow[0;\pi]$ and $\arctan:\mathbb{R}\rightarrow[-\pi/2;\pi/2]$, thus we should take into account $\arccos \circ \cos: y \in[\pi;2\pi] \mapsto 2\pi-y$, $\arctan \circ \tan: y \in[\pi/2;3\pi/2] \mapsto y - \pi$, and $\arctan \circ \tan: y \in[\pi/2;3\pi/2] \mapsto y - 2\pi$.
Thus
$$
\begin{aligned}
y \in [\pi;3\pi/2] \Rightarrow \cos^{-1}(\cos(y)) + \tan^{-1}(\tan(y)) = 2\pi - y + y - \pi = \pi,\\
y \in [3\pi/2;2\pi] \Rightarrow \cos^{-1}(\cos(y)) + \tan^{-1}(\tan(y)) = 2\pi - y + y - 2\pi = 0.\\
\end{aligned}
$$
Obviously, there are no solutions for $y \in [\pi;2\pi]$, but this result may be handy if you would ever like to plot the function.
The leftover is in essence equivalent to 
$$
\begin{aligned}
y \in [0;\pi/2] \Rightarrow y + y = \frac{2\pi}{3} \Rightarrow y &= \frac{\pi}{3},\\
y \in [\pi/2;\pi] \Rightarrow y + y - \pi = \frac{2\pi}{3} \Rightarrow y &= \frac{5\pi}{6}.\\
\end{aligned}
$$
Taking the overall period into account we get
$$
y = \frac{\pi}{3} + 2\pi n,\quad\text{and}\quad y = \frac{5\pi}{6} + 2\pi n.
$$
and you now only need to solve
$$
\left\{
\begin{aligned}
\frac{2x}{x^2-1} = \tan\left(\frac{\pi}{3} + 2\pi n\right) = \sqrt{3}\\
\frac{x^2-1}{x^2+1} = \cos\left(\frac{\pi}{3} + 2\pi n\right) = \frac{1}{2}
\end{aligned}
\right.
\quad\text{and}\quad
\left\{
\begin{aligned}
\frac{2x}{x^2-1} = \tan\left(\frac{5\pi}{6} + 2\pi n\right) = -\frac{1}{\sqrt{3}}\\
\frac{x^2-1}{x^2+1} = \cos\left(\frac{5\pi}{6} + 2\pi n\right) = -\frac{\sqrt{3}}{2}
\end{aligned}
\right.
$$
that are simple quadratic equations.
The result reads
$$
x = \sqrt{3}\quad\text{and}\quad x = 2-\sqrt{3}.
$$
A: Let $x=\tan\dfrac t2$ for $t\in(-\pi,\pi)$. The equation reads
$$\tan^{-1}(-\tan t)+\cos^{-1}(-\cos t)=\frac{2\pi}3$$
or
$$\tan^{-1}(\tan t)+\cos^{-1}(\cos  t)=\frac{\pi}3.$$
By lazily checking on a plot, we see that the function has a null slope in the negatives, where there are no solutions.
In the positives, the slope is $2$ and there are two roots:
$$2t=\frac\pi3$$ and $$2t-\pi=\frac\pi3.$$

You deduce $x$ from $t$.

$\tan\dfrac\pi3=\sqrt3$ or $\tan\dfrac\pi{12}=2-\sqrt3$.


Note:
Despite the use of a plot, this resolution is rigorous, because we know that these functions are piecewise linear and we know the limits of the pieces, so we could check formally.
A: Following your approach, you are letting $x=\tan(\theta)$ for some $\theta \in (-\pi/2,\pi/2)$, and arriving at $\frac{\pi}{3}=\arccos(\cos(2\theta))+\arctan(\tan(2\theta))$. The important thing is that $\arccos$ maps to $[0,\pi]$ while $\arctan$ maps to $(-\pi/2,\pi/2)$. Thus:


*

*if $2 \theta \not \in [0,\pi]$, then $\arccos(\cos(2\theta)) \neq 2\theta$. In this situation that occurs when $2\theta \in (-\pi,0)$, in which case the arccosine is $-2\theta$ (which geometrically is the corresponding point in the upper half plane).

*if $2\theta \not \in (-\pi/2,\pi/2)$ then $\arctan(\tan(2\theta)) \neq 2\theta$. This occurs here when $2\theta \in (-\pi,-\pi/2)$ and when $2\theta \in (\pi/2,\pi)$. In the first case you have a positive value of tangent in the third quadrant, so you need the corresponding point in the first quadrant, which is $2\theta+\pi$ (represented in $(-\pi/2,\pi/2)$). In the second case, you have a negative value of tangent in the second quadrant, so you need the corresponding point in the fourth quadrant (represented in $(-\pi/2,\pi/2)$). This is $2\theta-\pi$.


Thus you have two cases for the arccosine: $2\theta \in [0,\pi)$ in which case $\arccos(\cos(2\theta))=2\theta$ and $2\theta \in (-\pi,0]$ in which case $\arccos(\cos(2\theta))=-2\theta$. You have three cases for the arctangent: $2 \theta \in (-\pi/2,\pi/2)$ gives $\arctan(\tan(2\theta))=2\theta$, $2 \theta \in (-\pi,-\pi/2)$ gives $\arctan(\tan(2\theta))=2\theta+\pi$, and $2\theta \in (\pi/2,\pi)$ which gives $\arctan(\tan(2\theta))=2\theta-\pi$. Overall this results in four cases:


*

*$2\theta \in (-\pi,-\pi/2] \Rightarrow \arccos(\cos(2\theta))+\arctan(\tan(2\theta))=\pi$

*$2\theta \in (-\pi/2,0] \Rightarrow \arccos(\cos(2\theta))+\arctan(\tan(2\theta))=0$

*$2\theta \in [0,\pi/2) \Rightarrow \arccos(\cos(2\theta))+\arctan(\tan(2\theta))=4\theta$

*$2\theta \in [\pi/2,\pi) \Rightarrow \arccos(\cos(2\theta))+\arctan(\tan(2\theta))=4\theta-\pi$.


Case 3 gives $\theta=\pi/12$; case 4 gives $\theta=\pi/3$.
A: Another way:
From $\cos^{-1}\dfrac{1-x^2}{1+x^2}+\tan^{-1}\dfrac{2x}{1-x^2}=\dfrac\pi3$
Let $\cos^{-1}\dfrac{1-x^2}{1+x^2}=2y\implies0\le2y\le\pi$
$\cos2y=?,x^2=\tan^2y$
Case$\#1:$
If $x=\tan y\ge0$ 
$$\tan^{-1}(\tan2y)=\dfrac\pi3-2y\implies\tan(2y)=\tan\left(\dfrac\pi3-2y\right)$$
$\implies2y=n\pi+\dfrac\pi3-2y\iff y=\dfrac{\pi(3n+1)}{12}$ such that $0\le n<4$ and $\tan y\ge0$
$0\le \dfrac{\pi(3n+1)}{12}\le\dfrac\pi2\iff0\le3n+1\le6\implies n=0,1$
Case$\#2:$
If $x=-\tan y<0$
$$\tan^{-1}(\tan(-2y))=\dfrac\pi3-2y\implies\tan(-2y)=\tan\left(\dfrac\pi3-2y\right)$$
$\implies-2y=m\pi+\dfrac\pi3-2y\iff3m+1=0$ which is untenable as $m$ is any integer
