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:
[2, 6]
Using \(\cos 2x = 1-2\sin^2 x\) and letting \(s=\sin x\) (so \(s\in[-1,1]\)), the equation becomes \(1-2s^2+\alpha s = 2\alpha-7\).
Rearranging: \(8-2s^2+\alpha s-2\alpha=0\), i.e. \(\alpha(s-2) = 2s^2-8 = 2(s-2)(s+2)\).
Since \(s\in[-1,1]\), \(s-2\) is never zero, so dividing both sides gives \(\alpha = 2(s+2) = 2s+4\).
As \(s\) ranges over \([-1,1]\), \(\alpha=2s+4\) ranges over \([2(-1)+4,\ 2(1)+4] = [2,6]\).
So \(S=[2,6]\).
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