<s>[INST] 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? 
 [/INST] 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.  
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