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The relation sec x = sec α holds exactly when
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: A. x = 2nπ ± α, n ∈ Z
sec x = sec α is equivalent to cos x = cos α, so x = 2nπ ± α.
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