1. Given function

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

We need to determine where $f(x)$ is increasing or decreasing.

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2. Differentiate $f(x)$

First simplify the polynomial part:

$$\frac{4x^3-3x^2}{6}=\frac{2}{3}x^3-\frac{1}{2}x^2$$

Now differentiate term by term:

• $\frac{d}{dx}\left(\frac{2}{3}x^3\right)=2x^2$
• $\frac{d}{dx}\left(-\frac{1}{2}x^2\right)=-x$
• $\frac{d}{dx}(-2\sin x)=-2\cos x$
• For $(2x-1)\cos x$, use product rule: $$\frac{d}{dx}[(2x-1)\cos x]=2\cos x-(2x-1)\sin x$$

Therefore,

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

So,

$$f'(x)=2x^2-x-(2x-1)\sin x$$

Factor:

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

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3. Analyze the sign of $x-\sin x$

Consider the well-known inequality:

• For $x>0$, $\sin x < x \implies x-\sin x>0$
• For $x<0$, $\sin x > x \implies x-\sin x<0$
• At $x=0$, $x-\sin x=0$

Hence $x-\sin x$ has the same sign as $x$.

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4. Analyze the sign of $f'(x)$

Since

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

we check intervals separated by $x=0$ and $x=\frac12$.

(i) For $x<0$

• $2x-1<0$
• $x-\sin x<0$

So,

$$f'(x)>0$$

Hence $f$ is increasing on $(-\infty,0)$.

(ii) For $0<x<\frac12$

• $2x-1<0$
• $x-\sin x>0$

So,

$$f'(x)<0$$

Hence $f$ is decreasing on $(0,\tfrac12)$.

(iii) For $x>\frac12$

• $2x-1>0$
• $x-\sin x>0$

So,

$$f'(x)>0$$

Hence $f$ is increasing on $(\tfrac12,\infty)$.

Also,

$$f'\left(\frac12\right)=0$$

so the function is increasing on $\left[\frac12,\infty\right).$

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5. Check options

• A: increases in $\left(-\infty,\frac12\right]$ → false, because it decreases on $(0,\frac12)$.
• B: decreases in $\left(-\infty,\frac12\right]$ → false, because it increases on $(-\infty,0)$.
• C: increases in $\left[\frac12,\infty\right)$ → true.
• D: decreases in $\left[\frac12,\infty\right)$ → false.

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6. Final answer

The correct option is

$$\boxed{\text{C}}$$