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(A) Given the equation: $	an^{-1}(2x) + 	an^{-1}(3x) = \frac{\pi}{4}$ Applying the formula $	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{

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is a singleton set.

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(A) Given the equation: $	an^{-1}(2x) + 	an^{-1}(3x) = \frac{\pi}{4}$ Applying the formula $	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{1-AB} ight)$,we get: $	an^{-1} \left( \frac{2x + 3x}{1 - (2x)(3x)} ight) = \frac{\pi}{4}$ $\frac{5x}{1 - 6x^2} = 	an \left( \frac{\pi}{4} ight) = 1$ $5x = 1 - 6x^2$ $6x^2 + 5x - 1 = 0$ Factoring the quadratic equation: $(6x - 1)(x + 1) = 0$ This gives $x = \frac{1}{6}$ or $x = -1$. Since the condition is $x \geq 0$,we reject $x = -1$. Thus,the only solution is $x = \frac{1}{6}$. Therefore,the set contains only one element,which is a singleton set.

Considering only the principal values of the inverse trigonometric functions,the set $\{x \geq 0 : \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\}$

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