How do you solve #sin 2x=sinx # over the interval 0 to 2pi?

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Answer 1 · 2016-03-02T11:23:47

Solution is #x={0, pi/3, pi, (5pi)/3, 2pi}#

As #sin2x=2sinxcosx#,

#sin2x=sinx# is equivalent to

#2sinxcosx=sinx# or #2sinxcosx-sinx=0# or

#sinx(2cosx-1)=0# i.e.

either #sinx=0# and #x=0, pi, 2pi#

or #2cosx-1=0# i.e. #cosx=1/2# i.e. #x=pi/3, (5pi)/3#

Hence, Solution is #x={0, pi/3, pi, (5pi)/3, 2pi}#