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

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Answer 1 · 2015-02-17T17:38:11

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 $x=pi/3 and x=5/3pi$

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