Solution

Given,

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

Taking tangent on both sides,

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

Using,

$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

we get,

$\frac{3x+2x}{1-(3x)(2x)}=1$

$\frac{5x}{1-6x^2}=1$

So,

$5x=1-6x^2$

$6x^2+5x-1=0$

Factorizing,

$6x^2+6x-x-1=0$

$6x(x+1)-1(x+1)=0$

$(x+1)(6x-1)=0$

So,

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

Now check both values.

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

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

So, $x=\frac{1}{6}$ is valid.

For $x=-1$,

$\tan^{-1}(-3)+\tan^{-1}(-2)$ is negative, so it cannot be equal to $\frac{\pi}{4}$.

Therefore, only one solution exists.