The number of solutions of the equation 4 sin | vedclass

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Question

(D) Given equation: $4 \sin^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0$Substitute $\sin^2 x = 1 - \cos^2 x$:$4(1 - \cos^2 x) - 4 \cos^3 x + 9 - 4 \cos x = 0$

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$0$

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(D) Given equation: $4 \sin^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0$Substitute $\sin^2 x = 1 - \cos^2 x$:$4(1 - \cos^2 x) - 4 \cos^3 x + 9 - 4 \cos x = 0$$4 - 4 \cos^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0$$-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$Let $f(t) = 4t^3 + 4t^2 + 4t$ where $t = \cos x \in [-1, 1]$.The maximum value of $f(t)$ on $[-1, 1]$ occurs at $t = 1$:$f(1) = 4(1)^3 + 4(1)^2 + 4(1) = 4 + 4 + 4 = 12$.Since the maximum value of the left-hand side is $12$,it can never equal $13$.Therefore,there are no solutions.

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

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