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

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$[2, 6]$

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(B) Given equation: $\cos\, 2x + a\, \sin\, x = 2a - 7$Using $\cos\, 2x = 1 - 2\, \sin^2 x$,we get:$1 - 2\, \sin^2 x + a\, \sin\, x = 2a - 7$$2\, \sin^2 x - a\, \sin\, x + 2a - 8 = 0$This is a quadratic equation in $\sin\, x$. Solving for $\sin\, x$ using the quadratic formula:$\sin\, x = \frac{a \pm \sqrt{a^2 - 4(2)(2a - 8)}}{2(2)} = \frac{a \pm \sqrt{a^2 - 16a + 64}}{4}$$\sin\, x = \frac{a \pm \sqrt{(a - 8)^2}}{4} = \frac{a \pm (a - 8)}{4}$Two possible values for $\sin\, x$ are:$\sin\, x = \frac{a + a - 8}{4} = \frac{2a - 8}{4} = \frac{a - 4}{2}$$\sin\, x = \frac{a - a + 8}{4} = \frac{8}{4} = 2$Since $\sin\, x$ cannot be $2$,we must have $\sin\, x = \frac{a - 4}{2}$.For a solution to exist,$-1 \leq \sin\, x \leq 1$,so:$-1 \leq \frac{a - 4}{2} \leq 1$$-2 \leq a - 4 \leq 2$$2 \leq a \leq 6$Thus,the set of values for $a$ is $[2, 6]$.

The set of values of $a$ for which the equation $\cos\, 2x + a\, \sin\, x = 2a - 7$ possesses a solution is:

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