Let S be the set of all alpha in R such that | vedclass

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

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$[2, 6]$

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(C) Given equation: $\cos 2x + \alpha \sin x = 2\alpha - 7$.Using the identity $\cos 2x = 1 - 2\sin^2 x$,we get:$1 - 2\sin^2 x + \alpha \sin x = 2\alpha - 7$.Rearranging the terms to form a quadratic equation in $\sin x$:$2\sin^2 x - \alpha \sin x + (2\alpha - 8) = 0$.Using the quadratic formula $\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$\sin x = \frac{\alpha \pm \sqrt{\alpha^2 - 4(2)(2\alpha - 8)}}{4} = \frac{\alpha \pm \sqrt{\alpha^2 - 16\alpha + 64}}{4}$.Since $\alpha^2 - 16\alpha + 64 = (\alpha - 8)^2$,we have:$\sin x = \frac{\alpha \pm (\alpha - 8)}{4}$.This gives two possible values for $\sin x$:$1) \sin x = \frac{\alpha + \alpha - 8}{4} = \frac{2\alpha - 8}{4} = \frac{\alpha - 4}{2}$.$2) \sin x = \frac{\alpha - \alpha + 8}{4} = \frac{8}{4} = 2$.Since $\sin x = 2$ is impossible,we must have $\sin x = \frac{\alpha - 4}{2}$.For a solution to exist,$-1 \leq \sin x \leq 1$ must hold:$-1 \leq \frac{\alpha - 4}{2} \leq 1$.$-2 \leq \alpha - 4 \leq 2$.Adding $4$ to all parts: $2 \leq \alpha \leq 6$.Thus,$S = [2, 6]$.

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

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