(Tan^-1X+tan^-1 2x)=p/4,find the value of X.

Dear Student,

Assuming '(Tan^-1(x)+tan^-1 (2x)=pie/4'

$$\begin{gathered} ta{n}^{-1}\left(2x\right)+ta{n}^{-1}\left(x\right)=(\pi /4) ---\left(1\right) \\ So, u\sin g {\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\frac{(A+B)}{1-AB} \\ So \\ {\tan }^{-1}\left(2x\right)+{\tan }^{-1}\left(x\right)={\tan }^{-1}\frac{(2x+x)}{1-2x^2} \\ So, taking \tan on both sides we get \\ \frac{(2x+x)}{1-2x^2}=1 \\ 2x^2+3x-1=0 \\ U\sin g x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-3\pm \sqrt{9+8}}{4}=\frac{-3\pm \sqrt{17}}{4} \\ but x=\frac{-3-\sqrt{17}}{4} does not satisfy \left(1\right), \\ So, x=\frac{-3+\sqrt{17}}{4} \end{gathered}$$
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Regards!