1. We are given the equation $$2(\sin^6 x + \cos^6 x) - 3(\sin^4 x + \cos^4 x) + 1 = 0$$ and need to solve it. 2. Recall the Pythagorean identity: $$\sin^2 x + \cos^2 x = 1$$. 3. Use the formulas for powers of sine and cosine: - $$\sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3$$ - $$\sin^4 x + \cos^4 x = (\sin^2 x)^2 + (\cos^2 x)^2$$ 4. Let $$a = \sin^2 x$$ and $$b = \cos^2 x$$, so $$a + b = 1$$. 5. Use the sum of cubes formula: $$a^3 + b^3 = (a + b)(a^2 - ab + b^2) = 1 \cdot (a^2 - ab + b^2) = a^2 - ab + b^2$$ 6. Also, $$a^2 + b^2 = (a + b)^2 - 2ab = 1 - 2ab$$. 7. Substitute into the original equation: $$2(a^3 + b^3) - 3(a^2 + b^2) + 1 = 0$$ $$2(a^2 - ab + b^2) - 3(1 - 2ab) + 1 = 0$$ 8. Expand and simplify: $$2a^2 - 2ab + 2b^2 - 3 + 6ab + 1 = 0$$ $$2a^2 + 4ab + 2b^2 - 2 = 0$$ 9. Since $$a^2 + b^2 = 1 - 2ab$$, rewrite: $$2(1 - 2ab) + 4ab - 2 = 0$$ $$2 - 4ab + 4ab - 2 = 0$$ $$0 = 0$$ 10. The equation simplifies to an identity, meaning it holds for all $$x$$. **Final answer:** The equation is true for all real values of $$x$$.