SOLUTION: Solve the eq on the interval [0,2pi) : sin2x-cosx=0
I used the indetity sin2x = 2sinxcosx
So 2sinxcosx-cosx=0
Cosx(2sinx-1) = 0
Thus cosx=0 and 2sinx-1=0
Simplified i
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-> SOLUTION: Solve the eq on the interval [0,2pi) : sin2x-cosx=0
I used the indetity sin2x = 2sinxcosx
So 2sinxcosx-cosx=0
Cosx(2sinx-1) = 0
Thus cosx=0 and 2sinx-1=0
Simplified i
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Question 1121834: Solve the eq on the interval [0,2pi) : sin2x-cosx=0
I used the indetity sin2x = 2sinxcosx
So 2sinxcosx-cosx=0
Cosx(2sinx-1) = 0
Thus cosx=0 and 2sinx-1=0
Simplified im left with these to evaluate
Where is cosx = 0 and where is sinx= -1/2
Cosx = 0 at 3pi/2 and pi/2
Sinx = -1/2 at 7pi/ and 11pi/6
Would those be my values? Found 2 solutions by solver91311, Alan3354:Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website! Solve the eq on the interval [0,2pi) : sin2x-cosx=0
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Is it sin(2x)-cos(x) = 0 ?
Or sin^2(x) - cos(x) = 0 ?
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It's your job to make it clear.
