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? 

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?
