1. **State the problem:** Given the equation $\sin 2A = 2 \sin A$, find the value(s) of $A$. 2. **Recall the double-angle formula for sine:** $$\sin 2A = 2 \sin A \cos A$$ This formula expresses $\sin 2A$ in terms of $\sin A$ and $\cos A$. 3. **Substitute the formula into the given equation:** $$2 \sin A \cos A = 2 \sin A$$ 4. **Simplify the equation:** Divide both sides by 2: $$\sin A \cos A = \sin A$$ 5. **Rearrange the equation:** $$\sin A \cos A - \sin A = 0$$ Factor out $\sin A$: $$\sin A (\cos A - 1) = 0$$ 6. **Solve for $A$ using the zero product property:** Either $$\sin A = 0$$ or $$\cos A - 1 = 0 \implies \cos A = 1$$ 7. **Find solutions for $\sin A = 0$:** $$A = n\pi, \quad n \in \mathbb{Z}$$ 8. **Find solutions for $\cos A = 1$:** $$A = 2n\pi, \quad n \in \mathbb{Z}$$ 9. **Combine solutions:** Since $A = 2n\pi$ is a subset of $A = n\pi$, the complete solution is: $$A = n\pi, \quad n \in \mathbb{Z}$$ **Final answer:** $$A = n\pi, \quad n \in \mathbb{Z}$$