1. **State the problem:** Solve the equation $$\sin 5\theta + \sin 3\theta = 0$$ for $$0 \leq \theta \leq \pi$$. 2. **Use the sum-to-product formula:** Recall that $$\sin A + \sin B = 2 \sin \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$$. 3. **Apply the formula:** $$\sin 5\theta + \sin 3\theta = 2 \sin \left( \frac{5\theta + 3\theta}{2} \right) \cos \left( \frac{5\theta - 3\theta}{2} \right) = 2 \sin 4\theta \cos \theta$$ 4. **Set the product equal to zero:** $$2 \sin 4\theta \cos \theta = 0$$ 5. **Solve each factor separately:** - $$\sin 4\theta = 0$$ - $$\cos \theta = 0$$ 6. **Solve $$\sin 4\theta = 0$$:** $$\sin 4\theta = 0 \implies 4\theta = n\pi, \quad n \in \mathbb{Z}$$ Since $$0 \leq \theta \leq \pi$$, then $$0 \leq 4\theta \leq 4\pi$$. So $$n\pi \in [0,4\pi] \implies n = 0,1,2,3,4$$. Thus, $$\theta = \frac{n\pi}{4}, \quad n=0,1,2,3,4$$ 7. **Solve $$\cos \theta = 0$$:** $$\cos \theta = 0 \implies \theta = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$ Within $$0 \leq \theta \leq \pi$$, only $$\theta = \frac{\pi}{2}$$ fits. 8. **Combine all solutions:** $$\theta = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$$ Note that $$\theta=0$$ and $$\theta=\pi$$ are included since the inequality is $$\leq$$. **Final answer:** $$\boxed{\theta = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi}$$