1. State the problem: Solve the equation $\sin 2x - \sin x = 0$. 2. Use the double-angle formula: $\sin 2x = 2 \sin x \cos x$, so the equation becomes: $$2 \sin x \cos x - \sin x = 0$$ 3. Factor out $\sin x$: $$\sin x (2 \cos x - 1) = 0$$ 4. Set each factor equal to zero and solve separately: - $\sin x = 0$ - $2 \cos x - 1 = 0$ 5. Solve $\sin x = 0$: $$x = n\pi, \quad n \in \mathbb{Z}$$ 6. Solve $2 \cos x - 1 = 0$: $$2 \cos x = 1 \implies \cos x = \frac{1}{2}$$ 7. Find $x$ from $\cos x = \frac{1}{2}$: $$x = \pm \frac{\pi}{3} + 2 n \pi, \quad n \in \mathbb{Z}$$ 8. Final solution set: $$x = n\pi \quad \text{or} \quad x = \pm \frac{\pi}{3} + 2 n \pi, \quad n \in \mathbb{Z}$$ This means $x$ is any integer multiple of $\pi$, or $x$ is $\frac{\pi}{3}$ plus any integer multiple of $2\pi$, or $-\frac{\pi}{3}$ plus any integer multiple of $2\pi$.