1. **State the problem:** Solve the equation $$\tan 2x + 3 \tan x = 0$$ for $x$. 2. **Recall the double-angle formula for tangent:** $$\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}$$. 3. **Substitute the formula into the equation:** $$\frac{2 \tan x}{1 - \tan^2 x} + 3 \tan x = 0$$ 4. **Multiply both sides by the denominator $1 - \tan^2 x$ to clear the fraction:** $$2 \tan x + 3 \tan x (1 - \tan^2 x) = 0$$ 5. **Distribute $3 \tan x$:** $$2 \tan x + 3 \tan x - 3 \tan^3 x = 0$$ 6. **Combine like terms:** $$5 \tan x - 3 \tan^3 x = 0$$ 7. **Factor out $\tan x$:** $$\tan x (5 - 3 \tan^2 x) = 0$$ 8. **Set each factor equal to zero:** - Case 1: $$\tan x = 0$$ - Case 2: $$5 - 3 \tan^2 x = 0$$ 9. **Solve Case 1:** $$\tan x = 0 \implies x = n\pi, \quad n \in \mathbb{Z}$$ 10. **Solve Case 2:** $$5 - 3 \tan^2 x = 0 \implies 3 \tan^2 x = 5 \implies \tan^2 x = \frac{5}{3}$$ 11. **Take square root:** $$\tan x = \pm \sqrt{\frac{5}{3}}$$ 12. **General solution for tangent:** $$x = \arctan \left( \pm \sqrt{\frac{5}{3}} \right) + n\pi, \quad n \in \mathbb{Z}$$ **Final answer:** $$x = n\pi \quad \text{or} \quad x = \arctan \left( \sqrt{\frac{5}{3}} \right) + n\pi \quad \text{or} \quad x = \arctan \left( -\sqrt{\frac{5}{3}} \right) + n\pi, \quad n \in \mathbb{Z}$$