Solve the equation
​cos^-1(x²-1/x²+1) + tan^-1(2x/x²-1) =2π/3

Here , cos^-1 ( x² – 1 / x²  + 1) + tan^-1( 2x/ x² – 1)  = 2π/3
LHS = cos^-1 ( x² – 1 / x²  + 1) + tan^-1( 2x/ x² – 1) 
Dividing throughout by x² ,
cos^-1 (  1 – (1/x²) /( 1 + ( 1/x²) ) + tan^-1( 2/x / 1 – (1/x²) )
= cos^-1(  1   - (1/x)² / 1 +  (1/x)² ) + tan^-1 ( 2 (1/x)   / 1 – (1/x)² )
=2tan^-1(1/x) + 2tan^-1(1/x)    [ using 2tan^-1x = cos^-1( 1 – x² / 1 + x² ) = tan^-1( 2x / 1 – x²) ]
=4tan^-1(1/x)  
RHS = 2 π/3
So 4 tan^-1( 1/x )  = 2 π/3
Or tan^-1(1/x)   = π/6
Or 1/x = 1/√3
Or x = √3
Hope that helps.