1. **State the problem:** Find the solutions of the equation $$\sin 4\theta + 2 \sin \theta \cos \theta = 0.$$\n\n2. **Recall formulas:** Use the double-angle and multiple-angle identities:\n- $$\sin 4\theta = 2 \sin 2\theta \cos 2\theta$$\n- $$\sin 2\theta = 2 \sin \theta \cos \theta$$\nAlso, note that $$2 \sin \theta \cos \theta = \sin 2\theta.$$\n\n3. **Rewrite the equation:** Substitute these identities into the equation:\n$$\sin 4\theta + 2 \sin \theta \cos \theta = 2 \sin 2\theta \cos 2\theta + \sin 2\theta = 0.$$\n\n4. **Factor the expression:** Factor out $$\sin 2\theta$$:\n$$\sin 2\theta (2 \cos 2\theta + 1) = 0.$$\n\n5. **Solve each factor:**\n- Case 1: $$\sin 2\theta = 0$$\n- Case 2: $$2 \cos 2\theta + 1 = 0 \Rightarrow \cos 2\theta = -\frac{1}{2}.$$\n\n6. **Find solutions for Case 1:**\n$$\sin 2\theta = 0 \Rightarrow 2\theta = n\pi, \quad n \in \mathbb{Z}.$$\nTherefore, $$\theta = \frac{n\pi}{2}.$$\n\n7. **Find solutions for Case 2:**\n$$\cos 2\theta = -\frac{1}{2}.$$\nRecall that $$\cos x = -\frac{1}{2}$$ at $$x = \frac{2\pi}{3} + 2k\pi$$ and $$x = \frac{4\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}.$$\nSo,\n$$2\theta = \frac{2\pi}{3} + 2k\pi \quad \text{or} \quad 2\theta = \frac{4\pi}{3} + 2k\pi.$$\nDivide both sides by 2:\n$$\theta = \frac{\pi}{3} + k\pi \quad \text{or} \quad \theta = \frac{2\pi}{3} + k\pi.$$\n\n**Final solutions:**\n$$\boxed{\theta = \frac{n\pi}{2}, \quad \theta = \frac{\pi}{3} + k\pi, \quad \theta = \frac{2\pi}{3} + k\pi, \quad n,k \in \mathbb{Z}}.$$