# How do you solve #sin2x-cosx=0 # over the interval 0 to 2pi?

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May 9, 2016

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Use the identity

#2sinxcosx-cosx=0#

Factor a

#cosx(2sinx-1)=0#

Set both of these terms equal to

#cosx=0" "=>" "x=pi/2,(3pi)/2#

#2sinx-1=0" "=>" "sinx=1/2" "=>" "x=pi/6,(5pi)/6#

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