1. **Problem statement:** Solve the equation $$\cos 3x + \cos 2x + \cos x = 0$$ for $$0 \leq x \leq 2\pi$$. 2. **Formula and identities:** Recall the cosine addition formulas and sum-to-product identities. We can use the sum-to-product formula: $$\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$$ 3. **Apply sum-to-product:** Group $$\cos 3x + \cos x$$: $$\cos 3x + \cos x = 2 \cos \left( \frac{3x + x}{2} \right) \cos \left( \frac{3x - x}{2} \right) = 2 \cos 2x \cos x$$ 4. **Rewrite the equation:** Substitute back: $$2 \cos 2x \cos x + \cos 2x = 0$$ 5. **Factor out $$\cos 2x$$:** $$\cos 2x (2 \cos x + 1) = 0$$ 6. **Solve each factor:** - $$\cos 2x = 0$$ - $$2 \cos x + 1 = 0 \implies \cos x = -\frac{1}{2}$$ 7. **Solve $$\cos 2x = 0$$:** $$\cos 2x = 0 \implies 2x = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$ $$x = \frac{\pi}{4} + \frac{k\pi}{2}$$ Within $$0 \leq x \leq 2\pi$$, values of $$k$$ are 0,1,2,3: $$x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$ 8. **Solve $$\cos x = -\frac{1}{2}$$:** $$x = \frac{2\pi}{3}, \frac{4\pi}{3}$$ within $$0 \leq x \leq 2\pi$$ 9. **Final solution set:** $$x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{2\pi}{3}, \frac{4\pi}{3}$$ 10. **Verification:** These values satisfy the original equation and can be verified using Desmos by plotting $$y = \cos 3x + \cos 2x + \cos x$$ and checking zeros.