1. **Problem statement:** Solve the equation $$0 = \sin 2x - \cos x$$ for all values of $x$. 2. **Formula and identities:** Recall the double-angle identity for sine: $$\sin 2x = 2 \sin x \cos x$$ 3. **Rewrite the equation using the identity:** $$0 = 2 \sin x \cos x - \cos x$$ 4. **Factor the right-hand side:** $$0 = \cos x (2 \sin x - 1)$$ 5. **Set each factor equal to zero:** - Case 1: $$\cos x = 0$$ - Case 2: $$2 \sin x - 1 = 0$$ 6. **Solve Case 1:** $$\cos x = 0 \implies x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$ 7. **Solve Case 2:** $$2 \sin x - 1 = 0 \implies \sin x = \frac{1}{2}$$ 8. **Find $x$ for Case 2:** $$x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$ 9. **Final solution:** $$x = \frac{\pi}{2} + k\pi, \quad x = \frac{\pi}{6} + 2k\pi, \quad x = \frac{5\pi}{6} + 2k\pi, \quad k \in \mathbb{Z}$$