1. **State the problem:** Solve the equation $$\arctan(3x) + \arctan(2x) = \frac{\pi}{4}$$ for $x$. 2. **Formula used:** The sum of arctangents formula is $$\arctan a + \arctan b = \arctan \left( \frac{a+b}{1 - ab} \right)$$ provided the result is in the correct quadrant. 3. **Apply the formula:** Let $a = 3x$ and $b = 2x$, then $$\arctan(3x) + \arctan(2x) = \arctan \left( \frac{3x + 2x}{1 - (3x)(2x)} \right) = \arctan \left( \frac{5x}{1 - 6x^2} \right)$$ 4. **Set equal to $\pi/4$:** Since $\arctan \left( \frac{5x}{1 - 6x^2} \right) = \frac{\pi}{4}$, we have $$\frac{5x}{1 - 6x^2} = 1$$ because $\arctan 1 = \frac{\pi}{4}$. 5. **Solve the equation:** $$5x = 1 - 6x^2$$ Rearranged: $$6x^2 + 5x - 1 = 0$$ 6. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=6$, $b=5$, $c=-1$. Calculate the discriminant: $$\Delta = 5^2 - 4 \times 6 \times (-1) = 25 + 24 = 49$$ 7. **Find roots:** $$x = \frac{-5 \pm \sqrt{49}}{2 \times 6} = \frac{-5 \pm 7}{12}$$ Two solutions: - $$x = \frac{-5 + 7}{12} = \frac{2}{12} = \frac{1}{6}$$ - $$x = \frac{-5 - 7}{12} = \frac{-12}{12} = -1$$ 8. **Check domain restrictions:** The sum formula for arctan is valid if $1 - ab \neq 0$, here $1 - 6x^2 \neq 0$, so $x \neq \pm \frac{1}{\sqrt{6}}$. Both solutions satisfy this. 9. **Final answer:** $$\boxed{x = \frac{1}{6} \text{ or } x = -1}$$