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Solving sin²x = sin²α yields
Ax = 2nπ ± α, n ∈ Z
Bx = nπ + α, n ∈ Z
Cx = nπ + (−1)^n α, n ∈ Z
Dx = nπ ± α, n ∈ Z
Answer & Solution
Correct answer: D. x = nπ ± α, n ∈ Z
sin²x = sin²α means sin x = ± sin α, and the two cases together give x = nπ ± α.
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