1. **State the problem:** Simplify the expression $$-\cos^4 x + 4 \sin x - \frac{4}{3} \sin^3 x$$ and check if it equals $$\frac{1}{4} \cos^4 x$$. 2. **Recall identities:** Use the Pythagorean identity $$\sin^2 x + \cos^2 x = 1$$ to express powers of sine or cosine in terms of the other. 3. **Rewrite the expression:** The expression is $$-\cos^4 x + 4 \sin x - \frac{4}{3} \sin^3 x$$. 4. **Check if it can be simplified to $$\frac{1}{4} \cos^4 x$$:** Move all terms to one side: $$-\cos^4 x + 4 \sin x - \frac{4}{3} \sin^3 x - \frac{1}{4} \cos^4 x = 0$$ 5. **Combine like terms:** $$-\cos^4 x - \frac{1}{4} \cos^4 x = -\frac{5}{4} \cos^4 x$$ So the equation becomes: $$-\frac{5}{4} \cos^4 x + 4 \sin x - \frac{4}{3} \sin^3 x = 0$$ 6. **Rewrite $$\cos^4 x$$ as $$(1 - \sin^2 x)^2$$:** $$-\frac{5}{4} (1 - \sin^2 x)^2 + 4 \sin x - \frac{4}{3} \sin^3 x = 0$$ 7. **Expand $$(1 - \sin^2 x)^2$$:** $$1 - 2 \sin^2 x + \sin^4 x$$ 8. **Substitute and expand:** $$-\frac{5}{4} (1 - 2 \sin^2 x + \sin^4 x) + 4 \sin x - \frac{4}{3} \sin^3 x = 0$$ $$-\frac{5}{4} + \frac{5}{2} \sin^2 x - \frac{5}{4} \sin^4 x + 4 \sin x - \frac{4}{3} \sin^3 x = 0$$ 9. **Group terms:** $$-\frac{5}{4} + 4 \sin x + \frac{5}{2} \sin^2 x - \frac{4}{3} \sin^3 x - \frac{5}{4} \sin^4 x = 0$$ 10. **Conclusion:** This is a quartic polynomial in $$\sin x$$ and does not simplify to zero for all $$x$$. Therefore, the original expression is not equal to $$\frac{1}{4} \cos^4 x$$. **Final answer:** The expression $$-\cos^4 x + 4 \sin x - \frac{4}{3} \sin^3 x$$ cannot be simplified to $$\frac{1}{4} \cos^4 x$$.