1. **State the problem:** Solve the equation $$2 \cos x - \cos \frac{x}{2} = 1$$ for $$0 \leq x \leq 2\pi$$. 2. **Recall the formula and rules:** We are dealing with trigonometric functions involving $$\cos x$$ and $$\cos \frac{x}{2}$$. The goal is to find all $$x$$ in the given interval that satisfy the equation. 3. **Rewrite the equation:** $$2 \cos x - \cos \frac{x}{2} = 1$$ 4. **Use substitution:** Let $$t = \frac{x}{2}$$, so $$x = 2t$$. Then: $$2 \cos (2t) - \cos t = 1$$ 5. **Use double angle formula:** $$\cos (2t) = 2 \cos^2 t - 1$$ 6. **Substitute:** $$2 (2 \cos^2 t - 1) - \cos t = 1$$ 7. **Simplify:** $$4 \cos^2 t - 2 - \cos t = 1$$ 8. **Bring all terms to one side:** $$4 \cos^2 t - \cos t - 3 = 0$$ 9. **Let $$y = \cos t$$, then:** $$4 y^2 - y - 3 = 0$$ 10. **Solve quadratic equation:** Using quadratic formula: $$y = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 4 \cdot (-3)}}{2 \cdot 4} = \frac{1 \pm \sqrt{1 + 48}}{8} = \frac{1 \pm 7}{8}$$ 11. **Calculate roots:** - $$y_1 = \frac{1 + 7}{8} = 1$$ - $$y_2 = \frac{1 - 7}{8} = -\frac{3}{4}$$ 12. **Recall $$y = \cos t$$, so:** - For $$\cos t = 1$$, $$t = 0 + 2k\pi$$ - For $$\cos t = -\frac{3}{4}$$, $$t = \pm \arccos \left(-\frac{3}{4}\right) + 2k\pi$$ 13. **Recall $$t = \frac{x}{2}$$, so:** - $$x = 2t$$ - For $$\cos t = 1$$, $$x = 0 + 4k\pi$$ - For $$\cos t = -\frac{3}{4}$$, $$x = 2 \arccos \left(-\frac{3}{4}\right) + 4k\pi$$ and $$x = 2 (2\pi - \arccos \left(-\frac{3}{4}\right)) + 4k\pi$$ 14. **Find solutions in $$0 \leq x \leq 2\pi$$:** - For $$x = 0$$ (when $$k=0$$) - For $$x = 2 \arccos \left(-\frac{3}{4}\right)$$ - For $$x = 2 (2\pi - \arccos \left(-\frac{3}{4}\right))$$ 15. **Final answer:** $$x = 0, \quad x = 2 \arccos \left(-\frac{3}{4}\right), \quad x = 4\pi - 2 \arccos \left(-\frac{3}{4}\right)$$ These are the solutions in the interval $$0 \leq x \leq 2\pi$$. **Note:** Since $$4\pi - 2 \arccos \left(-\frac{3}{4}\right) > 2\pi$$, we only take the solutions within $$0$$ to $$2\pi$$, so the last solution is adjusted to $$2 (2\pi - \arccos (-\frac{3}{4}))$$ which simplifies to $$4\pi - 2 \arccos (-\frac{3}{4})$$ but must be checked if it lies within the interval. Since $$2\pi \approx 6.283$$ and $$\arccos (-\frac{3}{4}) \approx 2.418$$, then $$2 (2\pi - 2.418) = 2 (6.283 - 2.418) = 2 \times 3.865 = 7.73$$ which is greater than $$2\pi$$, so this solution is outside the interval. Therefore, the solutions in $$[0, 2\pi]$$ are: $$x = 0, \quad x = 2 \arccos \left(-\frac{3}{4}\right)$$