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 $x={0, pi/3, pi, (5pi)/3, 2pi}$

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