Find roots of equation $(x^2+1)\cdot \arccos\left(\frac{2x}{1+x^2}\right)+2x\cdot \mathrm{sgn}(x^2-1)=0$ 
Find roots of equation $(x^2+1)\cdot \arccos\left(\frac{2x}{1+x^2}\right)+2x\cdot \mathrm{sgn}(x^2-1)=0$

One root is $x=1$ (checking functions $\arccos$ and $\mathrm{sgn}$).
Second root is $x=0.442$.
How to find the second root?
How to check if this equation only has two roots? 
 A: First i would look at $|x|<1$, where $\mathrm{sgn}(x^2-1)=-1$
$$(x^2+1)\cdot \arccos(\frac{2x}{1+x^2})-2x=0$$
$$\arccos(\frac{2x}{1+x^2}) = +\frac{2x}{1+x^2}$$
Using a substitution $u =\frac{2x}{1+x^2}$. You get 
$$ \arccos(u) = u$$
$$ u = \arccos(u)$$
From here you could use fixed-point iteration, after checking the necessary conditions.
For $|x|>1$ the analysis is almost the same. But for this case it seems like there is no solution real to the equation.
A: There isn't a closed form for this equation. Numeric methods will give you a way to approximate the non-trivial root.
Rewrite it as $$\arccos\left(\frac{2x}{1+x^2}\right)=\mathrm{sgn}(1-x^2)\frac{2x}{1+x^2}$$
Let $x=\tan \theta$ for $\theta\in\left(-\frac{\pi}2,\frac{\pi}2\right)$. Then $$\frac{2x}{1+x^2}=\frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$$ (work it out.)
When $|x|<1$
When $|\theta|<\frac{\pi}{4}$ which coincides with when $|x|<1$, $\arccos(\sin(2\theta))=\frac{\pi}{2}-2\theta$. Also $1-x^2>0$ and your equation becomes:
$$\frac{\pi}{2}-2\theta = \sin(2\theta)$$
Since $f(\theta)=2\theta+\sin2\theta$ is increasing on $(-\pi/4,\pi/4)$, there is at most one solution. Since $f(0)=0$ and  $f(\pi/4)=\frac{\pi}{2}+1>\frac{\pi}{2}$, there is exactly one $\theta\in(0,\pi/4)$ with $f(\theta)=\frac{\pi}{2}$.
Case: $x>1$
When $x>1$, that is $\theta\in(\pi/4,\pi/2)$, then
$$\arccos(\sin(2\theta)) = 2\theta-\frac{\pi}{2}$$
So you are trying to solve:
$$\pi/2=2\theta-\sin(2\theta)$$
The right side is strictly increasing [the derivative is $2(1-\cos(2\theta))$] and equal to $\frac{\pi}{2}$ when $\theta=\frac{\pi}{4}$, so it has no solution when $\theta>\pi/4$.
When $x<-1$, that is $\theta\in(-\pi/2,-\pi/4)$:
$$\arccos(\sin(2\theta))=2\theta+\frac{3\pi}{2}$$
So you want:$$\frac{3\pi}{2} = 2(-\theta)+\sin2(-\theta)$$
That's not possible since the right side is bounded by $\pi+1<\frac{3\pi}{2}$.
Case: $|x|=1$
Handle the $x=\pm 1$ case by hand: $x=1$ is a solution, $x=-1$ is not.
Approximating the non-trivial solution via recursion
You have found all the solutions. The solution $|x|<1$ has $\theta=\arctan(x)$ solving:
$$\theta = \frac{\pi}{4}-\frac{1}{2}\sin2\theta=:g(\theta)$$
Then using iteration, since $|g'(\theta)|<1$ in $(0,\pi/4)$, we can try iteration:
$$\theta_0=0, \theta_{n+1}=g(\theta_n)$$
For example:
$$\theta_{33} \approx 0.415856, x_{33}=\tan\theta_{33}\approx 0.441611$$
is okay.
Approximating root with Newton
Letting $\alpha=2\theta$, you are trying to solve $$h(\alpha):=\alpha+\sin\alpha -\frac{\pi}2=0$$
Applying Newton's method, let $\alpha_0=0$ and let:
$$\alpha_{n+1}=\alpha_n - \frac{h(\alpha_n)}{h'(\alpha_n)} = \alpha_n-\frac{\alpha_n+\sin\alpha_n -\frac\pi2}{1+\cos\alpha_n}$$
This runs much faster - it yields fifteen digits in five iterations:
$$x_{5}=\tan(2\alpha_5) \approx 0.4416107917053284$$
(Note: Originally claimed this converged no faster. That was quite wrong.)
A: Taking $x=\tan \theta$ and noting that $2x/(1+x^2)=\sin 2\theta$ will simplify the equation to $$\pi/2-2\theta+\sin 2\theta\cdot\mathrm{sgn}(\tan^2\theta-1)=0\\\implies f(\theta)+g(\theta)=0$$ where $$f(\theta)=\pi/2-2\theta,\ g(\theta)=\sin 2\theta\cdot \mathrm{sgn}(\tan^2\theta-1)=\left\{\begin{array}
 & 
 \sin 2\theta & \mathrm{if}\ \theta\ge \pi/4\\
-\sin 2\theta & \mathrm{if}\ \theta< \pi/4
\end{array}
\right.$$ And then we can graphically find the solution by plotting the two functions.  
A: We can start by simplifying the sign function. We know that $x^2 - 1$ is negative between $-1 < x < 1$ and positive or zero otherwise (I'm assuming that's the definition you're using for it), so we have the following two equations:
$$(x^2 + 1) \cdot \arccos{\left(\frac{2x}{1 + x^2}\right)} - 2x = 0, -1 < x < 1$$
$$(x^2 + 1) \cdot \arccos{\left(\frac{2x}{1 + x^2}\right)} + 2x = 0, x < -1, x > 1$$
The first equation simplifies to :
$$ \arccos{\left(\frac{2x}{1 + x^2}\right)} = \frac{2x}{x^2 + 1}$$
We can do the division since $x^2 + 1$ is always positive. Now, let's use the substitution $u = \frac{2x}{x^2 + 1}$:
$$\arccos{(u)} = u$$
Which has the sole solution $u \approx 0.739085$. Back substituting u and solving for x gives $0.739085 = \frac{2x}{x^2 + 1}$ which leads to $x = 0.442$ (the one you mentioned) as well as $x = 2.264$.  
Now, noting that the second equation reduces to:
 $$ \arccos{\left(\frac{2x}{1 + x^2}\right)} = \frac{-2x}{x^2 + 1}$$
Can you use the same substitution and continue the process to find the roots?
