We have `tan^(-1)(2x)+ tan^(-1)(3x) = (pi)/(4)`
Then . `tan^(-1)((2x+3x)/(1-2x(3x))) = (pi)/(4)`
`rArr (5x)/(1-6x^(2)) tan ((pi)/(4))`
`rArr " "(5x)/(1-6x^(2)) = 1`
`rArr " "5x = 1 - 6x^(2)`
`rArr " "6x^(2) + 5x - 1 =0`
` rArr" "6x^(2)+6x - x- 1 = 0`
`rArr " " 6x(x+1) - (x+1) = 0`
`rArr " "(6x - 1) (x+1) = 0`
`rArr" "x = (1)/(6) or -1`.