Asked by dfg
Solve Sinx=Cos2x-1 for all values between 0 and 2pi
Answers
Answered by
Steve
cos 2x = 1 - 2sin^2 x
so, you have
2sin^2 x + sin x = 0
sin x (2sin x + 1) = 0
sinx = 0 --> x = 0,2pi
sinx = -1/2 --> x = 7pi/6,11pi/6
so, you have
2sin^2 x + sin x = 0
sin x (2sin x + 1) = 0
sinx = 0 --> x = 0,2pi
sinx = -1/2 --> x = 7pi/6,11pi/6
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