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Correct answer: 2pi - x
cos⁻¹(cos x) returns a value in [0, pi]. For x in [pi, 2pi], x itself is not in [0, pi] (except endpoints). Since cos(2pi - x) = cos x and (2pi - x) lies in [0, pi] for x in [pi, 2pi], we have cos⁻¹(cos x) = 2pi - x.
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