1. **State the problem:** Solve for $\theta$ in the equation $$4 - 4 \cos(2\theta) = -8 \sin(\theta)$$ where $0 \leq \theta < 2\pi$. 2. **Use trigonometric identities:** Recall that $\cos(2\theta) = 1 - 2\sin^2(\theta)$. 3. **Substitute the identity:** $$4 - 4(1 - 2\sin^2(\theta)) = -8 \sin(\theta)$$ 4. **Simplify the left side:** $$4 - 4 + 8 \sin^2(\theta) = -8 \sin(\theta)$$ which simplifies to $$8 \sin^2(\theta) = -8 \sin(\theta)$$ 5. **Divide both sides by 8:** $$\cancel{8} \sin^2(\theta) = -\cancel{8} \sin(\theta)$$ which gives $$\sin^2(\theta) = -\sin(\theta)$$ 6. **Rewrite the equation:** $$\sin^2(\theta) + \sin(\theta) = 0$$ 7. **Factor the equation:** $$\sin(\theta)(\sin(\theta) + 1) = 0$$ 8. **Set each factor equal to zero:** - $\sin(\theta) = 0$ - $\sin(\theta) + 1 = 0 \Rightarrow \sin(\theta) = -1$ 9. **Find solutions for $\sin(\theta) = 0$ in $[0, 2\pi)$:** $$\theta = 0, \pi$$ 10. **Find solutions for $\sin(\theta) = -1$ in $[0, 2\pi)$:** $$\theta = \frac{3\pi}{2}$$ **Final answer:** $$\theta = 0, \pi, \frac{3\pi}{2}$$