The number of solutions of the equation 4 sin 2 x − 4 cos 3 x + 9 − 4 cos x = 0 ; x ∈ − 2 π , 2 π is:

Given: $4{\sin }^{2}x-4{\cos }^{3}x+9-4\cos x=0$

$\Rightarrow 4-4{\cos }^{2}x-4{\cos }^{3}x+9-4\cos x=0$

$\Rightarrow 4{\cos }^{3}x+4{\cos }^{2}x+4\cos x-13=0$

$\Rightarrow 4{\cos }^{3}x+4{\cos }^{2}x+4\cos x=13$

We know that, $-1\le \cos x\le 1$

So, the maximum value of LHS is $\left(4+4+4\right)=12$

But, RHS is $13$.

So, no solutions are possible for the given equation.