1. **State the problem:** Solve the trigonometric equation $$\cos 2x - \cos x - 1 = 0$$. 2. **Recall the double-angle formula:** $$\cos 2x = 2\cos^2 x - 1$$. 3. **Substitute the formula into the equation:** $$2\cos^2 x - 1 - \cos x - 1 = 0$$ 4. **Simplify the equation:** $$2\cos^2 x - \cos x - 2 = 0$$ 5. **Let $$y = \cos x$$, then the equation becomes:** $$2y^2 - y - 2 = 0$$ 6. **Solve the quadratic equation using the quadratic formula:** $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 16}}{4} = \frac{1 \pm \sqrt{17}}{4}$$ 7. **Calculate the roots:** $$y_1 = \frac{1 + \sqrt{17}}{4} \approx 1.28$$ (not possible since $$\cos x$$ must be between -1 and 1) $$y_2 = \frac{1 - \sqrt{17}}{4} \approx -0.78$$ (valid) 8. **Find $$x$$ such that $$\cos x = -0.78$$:** $$x = \pm \arccos(-0.78) + 2k\pi, \quad k \in \mathbb{Z}$$ 9. **Final solution:** $$x = \pm \arccos(-0.78) + 2k\pi, \quad k \in \mathbb{Z}$$