If tan^-1 x + 2 cot^-1x =(2π)/3, then x =

$tan^{-1} x + 2 cot^{-1}x =\frac{2\pi}{3}$

$⇒ tan^{-1}x + 2\left(\frac{\pi}{2}-tan^{-1}x\right)=\frac{2\pi}{3}⇒ tan^{-1}x =\frac{\pi}{3}⇒x = \sqrt{3}$