The function $$f(x) = \frac{4x^3 - 3x^2}{6} - 2\sin x + (2x - 1)\cos x$$:

We have $$f(x) = \frac{4x^3 - 3x^2}{6} - 2\sin x + (2x - 1)\cos x$$.

Differentiating term by term: $$\frac{d}{dx}\left(\frac{4x^3 - 3x^2}{6}\right) = \frac{12x^2 - 6x}{6} = 2x^2 - x$$.

$$\frac{d}{dx}(-2\sin x) = -2\cos x$$.

$$\frac{d}{dx}((2x - 1)\cos x) = 2\cos x - (2x - 1)\sin x$$.

So $$f'(x) = 2x^2 - x - 2\cos x + 2\cos x - (2x - 1)\sin x = 2x^2 - x - (2x - 1)\sin x$$.

Factoring: $$f'(x) = x(2x - 1) - (2x - 1)\sin x = (2x - 1)(x - \sin x)$$.

We know that $$x - \sin x \geq 0$$ for $$x \geq 0$$ and $$x - \sin x \leq 0$$ for $$x \leq 0$$ (since the derivative of $$x - \sin x$$ is $$1 - \cos x \geq 0$$, so $$x - \sin x$$ is increasing with value 0 at $$x = 0$$).

For $$x > \frac{1}{2}$$: $$(2x - 1) > 0$$ and $$(x - \sin x) > 0$$, so $$f'(x) > 0$$. The function is increasing on $$\left(\frac{1}{2}, \infty\right)$$.

Hence, the correct answer is Option A.