The solution of tan^-12x+tan^-13x=(π)/4 is

\(\tan^{-1}2x+\tan^{-1}3x=\frac{\pi}{4}\)

$⇒\tan^{-1}(\frac{2x+3x}{1-6x^2})=\frac{\pi}{4}$,   $6x^2<1$

$5x=1-6x^2$

$⇒6x^2+5x-1=0$, $6x^2<1$

$6x^5+6x-x-1=0$

so $x=\frac{1}{6},-1$

$x=\frac{1}{6}$