1. **State the problem:** Solve the trigonometric equation $$4 - 4\cos(2\theta) = -8\sin(\theta)$$ for $\theta$. 2. **Recall formulas and identities:** - Double angle formula for cosine: $$\cos(2\theta) = 1 - 2\sin^2(\theta)$$ - We will use this to rewrite the equation in terms of $\sin(\theta)$. 3. **Substitute the double angle formula:** $$4 - 4(1 - 2\sin^2(\theta)) = -8\sin(\theta)$$ 4. **Simplify inside the parentheses:** $$4 - 4 + 8\sin^2(\theta) = -8\sin(\theta)$$ 5. **Simplify the left side:** $$8\sin^2(\theta) = -8\sin(\theta)$$ 6. **Bring all terms to one side:** $$8\sin^2(\theta) + 8\sin(\theta) = 0$$ 7. **Factor out common term:** $$8\sin(\theta)(\sin(\theta) + 1) = 0$$ 8. **Set each factor equal to zero:** - $$8\sin(\theta) = 0 \implies \sin(\theta) = 0$$ - $$\sin(\theta) + 1 = 0 \implies \sin(\theta) = -1$$ 9. **Solve for $\theta$:** - For $$\sin(\theta) = 0$$, solutions are $$\theta = n\pi$$ where $n$ is any integer. - For $$\sin(\theta) = -1$$, solution is $$\theta = \frac{3\pi}{2} + 2n\pi$$ where $n$ is any integer. **Final answer:** $$\theta = n\pi \quad \text{or} \quad \theta = \frac{3\pi}{2} + 2n\pi, \quad n \in \mathbb{Z}$$