How do you find all values of x such that #sin 2x = sin x# and #0 <= x <= 2pi#?

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Answer 1 · 2016-05-22T13:48:25

#x=npi# or #x=2npi+-pi/3#

As #sin2x=sinx#, we have

#2sinxcosx=sinx# or

#2sinxcosx-sinx=0#

#sinx(2cosx-1)=0#

i.e. either #sinx=0# which implies #x=npi#

or #2cosx-1=0# i.e. #cosx=1/2=cos(+-pi/3)#

which implies #x=2npi+-pi/3#