1. **State the problem:** Solve the equation $$\cos 2x + \cos x = 0$$ for $$x$$ in the interval $$[0, 2\pi)$$. 2. **Use the double-angle formula:** Recall that $$\cos 2x = 2\cos^2 x - 1$$. 3. **Rewrite the equation:** Substitute to get $$2\cos^2 x - 1 + \cos x = 0$$. 4. **Rearrange into a quadratic form:** $$2\cos^2 x + \cos x - 1 = 0$$. 5. **Let $$y = \cos x$$:** The equation becomes $$2y^2 + y - 1 = 0$$. 6. **Solve the quadratic equation:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a=2$$, $$b=1$$, $$c=-1$$. 7. **Calculate the discriminant:** $$\Delta = 1^2 - 4 \times 2 \times (-1) = 1 + 8 = 9$$. 8. **Find the roots:** $$y = \frac{-1 \pm \sqrt{9}}{2 \times 2} = \frac{-1 \pm 3}{4}$$. 9. **Roots are:** $$y_1 = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}$$ and $$y_2 = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$$. 10. **Find $$x$$ for each root:** - For $$\cos x = \frac{1}{2}$$, $$x = \frac{\pi}{3}$$ or $$x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$$. - For $$\cos x = -1$$, $$x = \pi$$. 11. **Final solutions in $$[0, 2\pi)$$:** $$x = \frac{\pi}{3}, \pi, \frac{5\pi}{3}$$. **Answer:** $$\boxed{\left\{ \frac{\pi}{3}, \pi, \frac{5\pi}{3} \right\}}$$