Question 868329: Solve cos2x+sinx=1in[0,2pi)?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Solve cos(2x)+sinx=1in[0,2pi)?
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cos(2x) = cos^2(x)-sin^2(x) = 1-2sin^2(x)
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Your equation:
1-2sin^2(x)+sin(x) = 1
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Rearrange:
2sin^2(x)-sin(x) = 0
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Factor:
sin(x)[2sin(x)-1] = 0
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sin(x) = 0 or sin(x) = 1/2
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x = 0 or pi or pi/6 or (5/6)pi
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Cheers,
Stan H.
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