1. **State the problem:** Solve the trigonometric equation $$\cos 2x - \cos x = 0$$ for $x$. 2. **Recall the double-angle formula:** $$\cos 2x = 2\cos^2 x - 1$$. 3. **Substitute the formula into the equation:** $$2\cos^2 x - 1 - \cos x = 0$$ 4. **Rewrite the equation:** $$2\cos^2 x - \cos x - 1 = 0$$ 5. **Let $y = \cos x$, then the equation becomes:** $$2y^2 - y - 1 = 0$$ 6. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}$$ 7. **Find the roots:** - $$y_1 = \frac{1 + 3}{4} = 1$$ - $$y_2 = \frac{1 - 3}{4} = -\frac{1}{2}$$ 8. **Recall $y = \cos x$, so:** - For $$\cos x = 1$$, solutions are $$x = 2k\pi$$ where $k$ is any integer. - For $$\cos x = -\frac{1}{2}$$, solutions are $$x = \pm \frac{2\pi}{3} + 2k\pi$$ where $k$ is any integer. **Final answer:** $$x = 2k\pi \quad \text{or} \quad x = \pm \frac{2\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$$