How do you solve the equation #sin2x-sinx=0# on the interval [0,2π]?

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Gió Share
Feb 17, 2015

You can start by writing:
#sin(2x)=2sin(x)cos(x)#
So you get:
#2sin(x)cos(x)-sin(x)=0#
#sin(x)*[2cos(x)-1]=0#

This holds when:

#sin(x)=0# and then #x=0 or x=pi or x=2pi#

or when:

#2cos(x)-1=0#
#cos(x)=1/2# and then #x=pi/3 and x=5/3pi#

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