1. **Problem:** Find the values of $x$ such that $\cos(2x) = \cos(x)$ and $0 \leq x \leq 2\pi$. 2. **Formula and rules:** Use the cosine double-angle identity and the property that $\cos A = \cos B$ implies $A = B + 2k\pi$ or $A = -B + 2k\pi$ for any integer $k$. 3. **Step 1:** Recall the double-angle formula: $$\cos(2x) = 2\cos^2(x) - 1$$ 4. **Step 2:** Set the equation: $$2\cos^2(x) - 1 = \cos(x)$$ 5. **Step 3:** Rearrange to form a quadratic in $\cos(x)$: $$2\cos^2(x) - \cos(x) - 1 = 0$$ 6. **Step 4:** Let $y = \cos(x)$, then: $$2y^2 - y - 1 = 0$$ 7. **Step 5:** Solve the quadratic equation using the quadratic formula: $$y = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 2 \times (-1)}}{2 \times 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4}$$ 8. **Step 6:** Find the roots: - $y = \frac{1 + 3}{4} = 1$ - $y = \frac{1 - 3}{4} = -\frac{1}{2}$ 9. **Step 7:** Find $x$ values for each root within $0 \leq x \leq 2\pi$: - For $\cos(x) = 1$, $x = 0$ - For $\cos(x) = -\frac{1}{2}$, $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$ 10. **Final answer:** $$x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}$$ These are the values of $x$ in the interval $[0, 2\pi]$ satisfying $\cos(2x) = \cos(x)$.