1. **State the problem:** Solve the equation $\tan^2 x + 3 \tan x = 0$ for $x$. 2. **Use substitution:** Let $t = \tan x$. The equation becomes: $$t^2 + 3t = 0$$ 3. **Factor the quadratic:** $$t(t + 3) = 0$$ 4. **Set each factor to zero:** - $t = 0$ - $t + 3 = 0 \implies t = -3$ 5. **Solve for $x$ using $\tan x = t$:** - For $\tan x = 0$, solutions are: $$x = n\pi, \quad n \in \mathbb{Z}$$ - For $\tan x = -3$, solutions are: $$x = \arctan(-3) + n\pi, \quad n \in \mathbb{Z}$$ 6. **Final answer:** $$x = n\pi \quad \text{or} \quad x = \arctan(-3) + n\pi, \quad n \in \mathbb{Z}$$ This means the solutions repeat every $\pi$ radians because the tangent function has period $\pi$.