If $1<x<\sqrt{2}$,the number of solutions of the equation ${\tan }^{-1} (x-1)+{\tan }^{-1} x+{\tan }^{-1} (x+1)={\tan }^{-1} 3x$ is

The given equation can be written as

${\tan }^{-1} (x-1)+{\tan }^{-1} (x+1)={\tan }^{-1} 3x-{\tan }^{-1} x$

$\begin{array}{l}\Rightarrow {\tan }^{-1} \frac{x-1+x+1}{1-(x-1)(x+1)}={\tan }^{-1} \frac{3x-x}{1+3x^2}\\ \Rightarrow \frac{2x}{2-x^2}=\frac{2x}{1+3x^2}\Rightarrow x+3x^3=2x-x^3\\ \Rightarrow 4x^3-x=0\Rightarrow x\left(4x^2-1\right)=0\\ \Rightarrow x=0,x=\pm 1/2\end{array}$

None of which satisfies $1<x<\sqrt{2}$