1. Given function

$f(x)=4\cos^3 x+3\sqrt{3}\cos^2 x-10$

We need the number of points of local maxima in $(0,2\pi)$.

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2. Differentiate $f(x)$

Let $u=\cos x$. Then

$f(x)=4u^3+3\sqrt{3}u^2-10$

So,

$f'(x)=\left(12\cos^2 x+6\sqrt{3}\cos x\right)(-\sin x)$

Hence,

$f'(x)=-6\sin x\cos x(2\cos x+\sqrt{3})$

Critical points occur when

$f'(x)=0$

So,

1. $\sin x=0$
2. $\cos x=0$
3. $2\cos x+\sqrt{3}=0$

In $(0,2\pi)$, these give:

• $\sin x=0 \Rightarrow x=\pi$

• $\cos x=0 \Rightarrow x=\dfrac{\pi}{2},\dfrac{3\pi}{2}$

• $2\cos x+\sqrt{3}=0 \Rightarrow \cos x=-\dfrac{\sqrt{3}}{2}$

  $x=\frac{5\pi}{6},\frac{7\pi}{6}$

So critical points are

$x=\frac{\pi}{2},\frac{5\pi}{6},\pi,\frac{7\pi}{6},\frac{3\pi}{2}$

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3. Sign analysis of $f'(x)$

We check the sign of

$f'(x)=-6\sin x\cos x(2\cos x+\sqrt{3})$

Split the interval using the critical points:

(i) $\left(0,\frac{\pi}{2}\right)$

Take $x=\frac{\pi}{3}$:

• $\sin x>0$
• $\cos x>0$
• $2\cos x+\sqrt{3}>0$

So $f'(x)<0$.

(ii) $\left(\frac{\pi}{2},\frac{5\pi}{6}\right)$

Take $x=\frac{2\pi}{3}$:

• $\sin x>0$
• $\cos x<0$
• $2\cos x+\sqrt{3}>0$

Thus $\sin x\cos x(2\cos x+\sqrt{3})<0$, hence

$f'(x)>0$

(iii) $\left(\frac{5\pi}{6},\pi\right)$

Take $x=\frac{11\pi}{12}$:

• $\sin x>0$
• $\cos x<0$
• $2\cos x+\sqrt{3}<0$

Product is positive, so

$f'(x)<0$

(iv) $\left(\pi,\frac{7\pi}{6}\right)$

Take $x=\frac{13\pi}{12}$:

• $\sin x<0$
• $\cos x<0$
• $2\cos x+\sqrt{3}<0$

Product is negative, so

$f'(x)>0$

(v) $\left(\frac{7\pi}{6},\frac{3\pi}{2}\right)$

Take $x=\frac{4\pi}{3}$:

• $\sin x<0$
• $\cos x<0$
• $2\cos x+\sqrt{3}>0$

Product is positive, so

$f'(x)<0$

(vi) $\left(\frac{3\pi}{2},2\pi\right)$

Take $x=\frac{11\pi}{6}$:

• $\sin x<0$
• $\cos x>0$
• $2\cos x+\sqrt{3}>0$

Product is negative, so

$f'(x)>0$

So the sign pattern is:

$-, +, -, +, -, +$

across the critical points

$\frac{\pi}{2},\frac{5\pi}{6},\pi,\frac{7\pi}{6},\frac{3\pi}{2}$

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4. Identify local maxima

A local maximum occurs where $f'(x)$ changes from positive to negative.

From the sign changes:

• At $x=\frac{\pi}{2}$: $- \to +$ → local minimum
• At $x=\frac{5\pi}{6}$: $+ \to -$ → local maximum
• At $x=\pi$: $- \to +$ → local minimum
• At $x=\frac{7\pi}{6}$: $+ \to -$ → local maximum
• At $x=\frac{3\pi}{2}$: $- \to +$ → local minimum

Thus, the number of points of local maxima is

$2$

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5. Compare with options

Option $D$ corresponds to $2$.

So the correct answer is:

$\boxed{2}$