Given equation is
\(tan^{-1}\left(\frac1{(1 +2x)}\right) + tan^{-1}\left(\frac1{(1 +4x)}\right) = tan^{-1} (\frac{2}{x^2})\)
⇒ \(tan^{-1}\left(\cfrac{\frac{1}{1+2x} + \frac1{1 +4x}}{1 - \frac{1}{1+2x} \times\frac1{1+4x}}\right) = tan{-1}\left(\frac{2}{x^2}\right)\)
⇒ \(\frac{1 +4x + 1 +2x}{1 + 6x + 8x62 -1} = \frac2 {x^2}\)
⇒ \(\frac{2 + 6x}{6x + 8x^2} = \frac2{x^2}\)
⇒ \(\frac{1 +3x}{3x +4x^2} = \frac2{x^2}\)
⇒ 3x3 + x2 - 8x2 - 6x = 0
⇒ 3x3 - 7x2 - 6x = 0
⇒ x(3x2 - 7x - 6) = 0
⇒ x(3x2 - 9x + 2x -6) =0
⇒ x(3x(x - 3) + 2(x - 3)) = 0
⇒ x (x - 3) (3x + 2) = 0
⇒ x = 0 or x - 3 = 0 or 3x + 2 = 0
⇒ x =0 or x = 3 or x = \(\frac{-2}3\)
Possible value of x can be 0, 3 & \(\frac{-2}3\).