1. **State the problem:** Solve the equation $\sin 2x = 2 \sin x$. 2. **Recall the double-angle formula:** $\sin 2x = 2 \sin x \cos x$. 3. **Rewrite the equation using the formula:** $$2 \sin x \cos x = 2 \sin x$$ 4. **Bring all terms to one side:** $$2 \sin x \cos x - 2 \sin x = 0$$ 5. **Factor out the common term $2 \sin x$:** $$2 \sin x (\cos x - 1) = 0$$ 6. **Set each factor equal to zero:** - $2 \sin x = 0 \implies \sin x = 0$ - $\cos x - 1 = 0 \implies \cos x = 1$ 7. **Solve each equation:** - $\sin x = 0$ at $x = n\pi$, where $n$ is any integer. - $\cos x = 1$ at $x = 2n\pi$, where $n$ is any integer. 8. **Combine solutions:** Since $x = 2n\pi$ is included in $x = n\pi$, the full solution set is: $$x = n\pi, \quad n \in \mathbb{Z}$$ **Final answer:** $$x = n\pi, \quad n \in \mathbb{Z}$$