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Solution set of sin 2x = -1/2 in [0,2pi)
Answered by Steve
since sin(π/6) = 1/2
and sine is negative in QII and QIV,
2x = 7π/6 or 11π/6 + 2nπ
x = 7π/12 + nπ or 11π/12 + nπ
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