Given the inverse trigonometric equation: \[ \tan^{-1}(3x) + \tan^{-1}(2x) = \frac{\pi}{4} \] We apply the formula: \[ \tan^{-1}(A) + \tan^{-1}(B) = \tan^{-1}\left(\frac{A+B}{1-AB}\right) \] (Condition: \( AB < 1 \) for the standard principal branch result to be positive as needed here). Substituting \( A = 3x \) and \( B = 2x \): \[ \tan^{-1}\left(\frac{3x + 2x}{1 - (3x)(2x)}\right) = \frac{\pi}{4} \] \[ \frac{5x}{1 - 6x^2} = \tan\left(\frac{\pi}{4}\right) \] Since \( \tan\left(\frac{\pi}{4}\right) = 1 \): \[ \frac{5x}{1 - 6x^2} = 1 \] Rearranging into a quadratic equation: \[ 5x = 1 - 6x^2 \] \[ 6x^2 + 5x - 1 = 0 \] Factorizing the quadratic equation: \[ 6x^2 + 6x - x - 1 = 0 \] \[ 6x(x + 1) - 1(x + 1) = 0 \] \[ (6x - 1)(x + 1) = 0 \] This gives two possible values for \( x \): \[ x = \frac{1}{6} \quad \text{and} \quad x = -1 \] Now, we must verify the solutions against the original equation. Case 1: \( x = -1 \) \[ \tan^{-1}(-3) + \tan^{-1}(-2) \] Since both arguments are negative, their arctangents are negative angles in the 4th quadrant \( (-\pi/2, 0) \). The sum of two negative angles cannot equal \( +\frac{\pi}{4} \). Thus, \( x = -1 \) is an extraneous solution. Case 2: \( x = \frac{1}{6} \) \[ \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{3}\right) \] Both arguments are positive, giving angles in the 1st quadrant. Also, \( 3x \cdot 2x = 6\left(\frac{1}{6}\right)^2 = \frac{1}{6} < 1 \), meaning the identity holds without a \( \pi \) shift. This solution perfectly balances the equation. Therefore, there is exactly 1 valid solution.