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 $x=2npi+-pi/3$

How do you find all values of x such that sin 2x = sin xsin 2x = sin x and 0 <= x <= 2pi0 <= x <= 2pi?