The number of solutions of the equation 4 ^2 x-4 ^3 x+9-4 x=0 ; x [-2 , 2 ] is :

Question

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 :

Accepted Answer

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:

$$
4(1 -
\cos^2 x) - 4
\cos^3 x + 9 - 4
\cos x = 0
$$

Rearranging and simplifying this equation, we have:

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

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

$4 \cos ^3 x+4 \cos ^2 x+4 \cos x=13$

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.

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.

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