The provided equation is \(\tan^{-1}\left(\frac{2}{3-x+1}\right) = \cot^{-1}\left(\frac{3}{3x+1}\right)\).
Using the identity \(\tan^{-1}(y) + \cot^{-1}(y) = \frac{\pi}{2}\), the equation transforms to:
\[\tan^{-1}\left(\frac{2}{4-x}\right) + \tan^{-1}\left(\frac{1}{\frac{3}{3x+1}}\right) = \frac{\pi}{2}\]
Simplifying the second term yields:
\(\tan^{-1}\left(\frac{3x+1}{3}\right)\)
The equation is now:
\[\tan^{-1}\left(\frac{2}{4-x}\right) + \tan^{-1}\left(\frac{3x+1}{3}\right) = \frac{\pi}{2}\]
Applying the tangent addition formula \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\) implies:
\[\frac{\frac{2}{4-x} + \frac{3x+1}{3}}{1 - \left(\frac{2}{4-x}\right)\left(\frac{3x+1}{3}\right)} = \tan\left(\frac{\pi}{2}\right)\]
Since \(\tan\left(\frac{\pi}{2}\right)\) is undefined, the denominator must be zero:
\[1 - \left(\frac{2}{4-x}\right)\left(\frac{3x+1}{3}\right) = 0\]
\[1 = \left(\frac{2}{4-x}\right)\left(\frac{3x+1}{3}\right)\]
\[\frac{2(3x+1)}{3(4-x)} = 1\]
Eliminating fractions by cross-multiplication:
\[6x + 2 = 12 - 3x\]
Combining like terms:
\[9x = 10\]
\[x = \frac{10}{9}\]
To verify potential roots, consider the case where \(\frac{3x+1}{3} = -\frac{2}{4-x}\), which arises from a specific interpretation of the arctangent addition.
\[\frac{3x+1}{3} = -\frac{2}{4-x}\]
Cross-multiplying yields:
\[3(4-x) + 2(3x+1) = 0\]
\[12 - 3x + 6x + 2 = 0\]
Simplifying gives:
\[3x = -14\]
\[x = -\frac{14}{3}\]
Therefore, there are two distinct real solutions for \(x\), one positive and one negative.