Let ‘S’ be the set of all α ∈ R such that the equation cos 2x + α sin x = 2α – 7 has a solution. Then S is equal to:

From question, the equation given is:

cos 2x + α sin x = 2α – 7

∵ [cos 2x = 1 – 2 sin² x]

⇒ 1 – 2 sin² x + α sin x – 2α + 7 = 0

⇒ -2 sin² x + α sin x – 2α + 8 = 0

⇒ 2 sin²x – αsinx + 2α – 8 = 0

⇒ (2 sin² x – 8) – α sin x + 2α = 0

⇒ 2(sin² x – 4) – α(sin x – 2) = 0

⇒ 2(sin x – 2)(sin x + 2) – α (sin x – 2) = 0

⇒(sin x – 2)[2(sin x + 2) – α] = 0

Now, (sin x – 2):

⇒ sin x – 2 = 0

∴ sin x = 2

Which is not possible.

Now, 2(sin x + 2) – α:

⇒ 2(sin x + 2) – α = 0

⇒ (2 sin x + 4) = α

⇒ 2 sin x = α – 4

\(\therefore \sin x = \frac{{\alpha - 4}}{2}\)

Now, the range of the sin x is generally from -1 to 1

⇒ -1 < x < 1

\(\Rightarrow - 1 < \left( {\frac{{\alpha - 4}}{2}} \right) < 1\)

⇒ -2 < (α – 4) < 2

∴ 2 < α < 6