1. **State the problem:** Solve the equation $$\frac{\cos(2x)}{3} = \frac{1}{6}$$ for $$0 < x < 2\pi$$. 2. **Rewrite the equation:** Multiply both sides by 3 to isolate $$\cos(2x)$$: $$\cos(2x) = \frac{1}{6} \times 3 = \frac{1}{2}$$ 3. **Recall the cosine values:** We need to find $$2x$$ such that $$\cos(2x) = \frac{1}{2}$$. 4. **General solutions for cosine:** $$\cos(\theta) = \frac{1}{2} \implies \theta = \pm \frac{\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$$ 5. **Apply to $$2x$$:** $$2x = \frac{\pi}{3} + 2k\pi \quad \text{or} \quad 2x = -\frac{\pi}{3} + 2k\pi$$ 6. **Solve for $$x$$:** $$x = \frac{\pi}{6} + k\pi \quad \text{or} \quad x = -\frac{\pi}{6} + k\pi$$ 7. **Find all $$x$$ in $$0 < x < 2\pi$$:** - For $$x = \frac{\pi}{6} + k\pi$$: - $$k=0 \Rightarrow x=\frac{\pi}{6}$$ - $$k=1 \Rightarrow x=\frac{\pi}{6} + \pi = \frac{7\pi}{6}$$ - For $$x = -\frac{\pi}{6} + k\pi$$: - $$k=1 \Rightarrow x= -\frac{\pi}{6} + \pi = \frac{5\pi}{6}$$ - $$k=2 \Rightarrow x= -\frac{\pi}{6} + 2\pi = \frac{11\pi}{6}$$ 8. **List all solutions:** $$x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$$ 9. **Match with options:** This corresponds to Option 4. **Final answer:** $$x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$$