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(C) Given equation: $\cos 2x + \alpha \sin x = 2\alpha - 7$Using $\cos 2x = 1 - 2 \sin^2 x$,we get:$1 - 2 \sin^2 x + \alpha \sin x = 2\alpha - 7$$2 \s

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$[2, 6]$

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(C) Given equation: $\cos 2x + \alpha \sin x = 2\alpha - 7$Using $\cos 2x = 1 - 2 \sin^2 x$,we get:$1 - 2 \sin^2 x + \alpha \sin x = 2\alpha - 7$$2 \sin^2 x - \alpha \sin x + 2\alpha - 8 = 0$This is a quadratic equation in $\sin x$. Using the quadratic formula $\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$\sin x = \frac{\alpha \pm \sqrt{\alpha^2 - 4(2)(2\alpha - 8)}}{4}$$\sin x = \frac{\alpha \pm \sqrt{\alpha^2 - 16\alpha + 64}}{4}$$\sin x = \frac{\alpha \pm \sqrt{(\alpha - 8)^2}}{4}$$\sin x = \frac{\alpha \pm (\alpha - 8)}{4}$Case $1$: $\sin x = \frac{\alpha + \alpha - 8}{4} = \frac{2\alpha - 8}{4} = \frac{\alpha - 4}{2}$Case $2$: $\sin x = \frac{\alpha - (\alpha - 8)}{4} = \frac{8}{4} = 2$ (Not possible as $\sin x \in [-1, 1]$)For a solution to exist,we must have $-1 \leq \frac{\alpha - 4}{2} \leq 1$$-2 \leq \alpha - 4 \leq 2$$2 \leq \alpha \leq 6$Thus,$S = [2, 6]$.

Let $S$ be the set of all $\alpha \in \mathbb{R}$ such that the equation $\cos 2x + \alpha \sin x = 2\alpha - 7$ has a solution. Then $S$ is equal to

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