Question

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

Answer 1

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`