1. Given equation

We need to solve $4\sin^2 x - 4\cos^3 x + 9 - 4\cos x = 0, \qquad x\in[-2\pi,2\pi].$

We are asked for the number of solutions in the interval.

2. Use the identity $\sin^2 x = 1 - \cos^2 x$

Substitute into the equation: $4(1-\cos^2 x) - 4\cos^3 x + 9 - 4\cos x = 0.$

Expand: $4 - 4\cos^2 x - 4\cos^3 x + 9 - 4\cos x = 0.$

So, $13 - 4\cos x - 4\cos^2 x - 4\cos^3 x = 0.$

Rearrange: $4\cos^3 x + 4\cos^2 x + 4\cos x - 13 = 0.$

Let $t = \cos x, \qquad t\in[-1,1].$

Then we get the cubic equation $4t^3 + 4t^2 + 4t - 13 = 0.$

3. Check whether this cubic has any root in $[-1,1]$

Define $f(t)=4t^3+4t^2+4t-13.$

Now evaluate at the endpoints: $f(-1)=4(-1)^3+4(-1)^2+4(-1)-13=-4+4-4-13=-17,$ $f(1)=4+4+4-13=-1.$

Thus, $f(-1)=-17<0, \qquad f(1)=-1<0.$

So at both ends of the interval $[-1,1]$, the value is negative.

4. Check monotonicity on $[-1,1]$

Differentiate: $f'(t)=12t^2+8t+4=4(3t^2+2t+1).$

Now, $3t^2+2t+1$ has discriminant $\Delta = 2^2 - 4\cdot 3 \cdot 1 = 4-12 = -8 < 0.$

Since the leading coefficient is positive, we have $3t^2+2t+1>0 \quad \text{for all real } t.$

Hence, $f'(t)>0 \quad \text{for all real } t.$

So $f(t)$ is strictly increasing everywhere.

Because $f$ is increasing and still $f(1)=-1<0,$ we get $f(t)<0 \quad \text{for all } t\in[-1,1].$

Therefore, the equation $4t^3+4t^2+4t-13=0$ has no solution with $t=\cos x\in[-1,1]$.

5. Conclusion for $x$

Since $\cos x$ can only lie in $[-1,1]$, and there is no admissible $t$ in this interval satisfying the cubic, the original trigonometric equation has $\boxed{0}$ solutions in $[-2\pi,2\pi]$.

6. Option check

• A: $0$ ✅
• B: $3$ ❌
• C: $1$ ❌
• D: $2$ ❌

So the correct option is $\boxed{\text{A}}.$

7. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.