The number of solutions of the equation $4 \sin ^2 x-4 \cos ^3 x+9-4 \cos x=0 ; x \in[-2 \pi, 2 \pi]$ is :

<p>We start by recognizing that $ \sin^2 x + \cos^2 x = 1$. Substituting $ \sin^2 x = 1 - \cos^2 x$ into the original equation gives:</p> <p>$4(1 - \cos^2 x) - 4 \cos^3 x + 9 - 4 \cos x = 0$</p> <p>Rearranging and simplifying this equation, we have:</p> <p>$4 - 4 \cos^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0$</p> <p>$4 \cos^3 x + 4 \cos^2 x + 4 \cos x - 13 = 0$</p> <p>$4 \cos ^3 x+4 \cos ^2 x+4 \cos x=13$</p> <p>Observing the bounds given, $x \in [-2\pi, 2\pi]$, we want to find how many solutions satisfy this cubic equation in terms of $ \cos x$. However, we note that the left-hand side (LHS) of the equation, representing a combination of cosines, could at most approach a maximum sum when $ \cos x = 1$, that being $4(1) + 4(1) + 4(1) = 12$. Yet, we have the equation set to equal 13, which is impossible given the maximum sum of the LHS can only be 12.</p> <p>The above reasoning indicates that, within the domain specified, there is no value of $x$ for which the equation holds true. Therefore, the number of solutions to the equation is zero.</p>