1. **State the problem:** Prove that $$\sqrt{2} + \sqrt{2} + \cos 4\theta = 2 \cos \theta$$. 2. **Simplify the left side:** Note that $$\sqrt{2} + \sqrt{2} = 2\sqrt{2}$$, so the expression becomes $$2\sqrt{2} + \cos 4\theta$$. 3. **Check the problem statement:** The given equation $$2\sqrt{2} + \cos 4\theta = 2 \cos \theta$$ does not hold for all $$\theta$$ because the left side is always greater than or equal to $$2\sqrt{2} - 1$$ (since $$\cos 4\theta$$ ranges from -1 to 1), while the right side ranges from -2 to 2. 4. **Possible correction:** Perhaps the problem intended to prove a trigonometric identity involving $$\cos 4\theta$$ and $$\cos \theta$$ without the $$\sqrt{2}$$ terms. 5. **Common identity:** $$\cos 4\theta = 8 \cos^4 \theta - 8 \cos^2 \theta + 1$$ or $$\cos 4\theta = 2 (2 \cos^2 2\theta - 1)^2 - 1$$. 6. **Conclusion:** The original equation as stated is not a standard identity and cannot be proven true for all $$\theta$$. **Final answer:** The equation $$\sqrt{2} + \sqrt{2} + \cos 4\theta = 2 \cos \theta$$ is not generally true and cannot be proven as an identity.