1. **State the problem:** Solve the equation $$3\sin^2\theta + 9\sin\theta = 0$$ for $$0 \leq \theta < 2\pi$$. 2. **Rewrite the equation:** Let $$x = \sin\theta$$. The equation becomes $$3x^2 + 9x = 0$$. 3. **Factor the quadratic:** $$3x^2 + 9x = 3x(x + 3) = 0$$. 4. **Set each factor to zero:** - $$3x = 0 \implies x = 0$$ - $$x + 3 = 0 \implies x = -3$$ 5. **Check the domain of $$\sin\theta$$:** Since $$\sin\theta$$ ranges from $$-1$$ to $$1$$, $$x = -3$$ is not possible. 6. **Solve for $$\theta$$ when $$\sin\theta = 0$$:** - $$\sin\theta = 0$$ at $$\theta = 0, \pi, 2\pi$$. 7. **Check the interval $$0 \leq \theta < 2\pi$$:** - $$\theta = 0$$ and $$\theta = \pi$$ are included. - $$\theta = 2\pi$$ is excluded because the interval is $$[0, 2\pi)$$. **Final answer:** $$\theta = 0, \pi$$.