Number of solutions of the equation ${\tan }^{-1}\left(\frac{1}{2\mathrm{x}+1}\right)+{\tan }^{-1}\left(\frac{1}{4\mathrm{x}+1}\right)={\tan }^{-1}\left(\frac{2}{{\mathrm{x}}^2}\right)$ is

$\begin{array}{l}\text{ Given, }{\tan }^{-1}\left(\frac{1}{2\mathrm{x}+1}\right)+{\tan }^{-1}\left(\frac{1}{4\mathrm{x}+1}\right)={\tan }^{-1}\left(\frac{2}{{\mathrm{x}}^2}\right)\\ \Rightarrow {\tan }^{-1}\left(\frac{\frac{1}{2\mathrm{x}+1}+\frac{1}{4\mathrm{x}+1}}{1-\frac{1}{2\mathrm{x}+1}\times \frac{1}{4\mathrm{x}+1}}\right)={\tan }^{-1}\left(\frac{2}{{\mathrm{x}}^2}\right)\\ \Rightarrow {\tan }^{-1}\left(\frac{6\mathrm{x}+2}{8{\mathrm{x}}^2+6\mathrm{x}}\right)={\tan }^{-1}\left(\frac{2}{{\mathrm{x}}^2}\right)\\ \Rightarrow \frac{6\mathrm{x}+2}{8{\mathrm{x}}^2+6\mathrm{x}}=\frac{2}{{\mathrm{x}}^2}\\ \Rightarrow 6{\mathrm{x}}^3+2{\mathrm{x}}^2=16{\mathrm{x}}^2+12\mathrm{x}\\ \Rightarrow 6{\mathrm{x}}^3-14{\mathrm{x}}^2-12\mathrm{x}=0\\ \Rightarrow 2\mathrm{x}\left(3{\mathrm{x}}^2-7\mathrm{x}-6\right)=0\\ \Rightarrow 2\mathrm{x}(3\mathrm{x}+2)(\mathrm{x}-3)=0\\ \Rightarrow \mathrm{x}=0,-\frac{2}{3},3\end{array}$

But x= - ⅔, not satisfy the given relation.