1. **State the problem:** Solve the trigonometric equation $$\sin^6\theta + \cos^6\theta + 3\sin^2\theta \cos^2\theta = 1.$$ 2. **Recall useful identities:** We know that $$\sin^2\theta + \cos^2\theta = 1.$$ Also, the expression $$\sin^6\theta + \cos^6\theta$$ can be rewritten using the identity for sum of cubes: $$a^3 + b^3 = (a+b)^3 - 3ab(a+b)$$ where here $$a = \sin^2\theta$$ and $$b = \cos^2\theta$$. 3. **Rewrite $$\sin^6\theta + \cos^6\theta$$:** $$\sin^6\theta + \cos^6\theta = (\sin^2\theta)^3 + (\cos^2\theta)^3 = (\sin^2\theta + \cos^2\theta)^3 - 3\sin^2\theta \cos^2\theta (\sin^2\theta + \cos^2\theta).$$ 4. **Simplify using $$\sin^2\theta + \cos^2\theta = 1$$:** $$\sin^6\theta + \cos^6\theta = 1^3 - 3\sin^2\theta \cos^2\theta \cdot 1 = 1 - 3\sin^2\theta \cos^2\theta.$$ 5. **Substitute back into the original equation:** $$1 - 3\sin^2\theta \cos^2\theta + 3\sin^2\theta \cos^2\theta = 1.$$ 6. **Simplify the equation:** The terms $$-3\sin^2\theta \cos^2\theta$$ and $$+3\sin^2\theta \cos^2\theta$$ cancel out, leaving: $$1 = 1,$$ which is always true. 7. **Conclusion:** The equation holds for all values of $$\theta$$. Therefore, the solution is: $$\boxed{\text{All real } \theta \text{ satisfy the equation.}}$$