I need a little help on this one

[kaebun]

solve all solutions from [0,2pi)
sin2x=sinx so i did
2sinxcosx=sinx
i don't know what else to do... am just not imaginitive or what? i just can't think of anything that will work

 

[Unco]

Your imagination is just fine, Kaebun.

Now you can move the sin(x) to the LHS, factorise, and solve just as we normally would.

 

[kaebun]

so ...
sinx(2cosx-1)=0
sinx=0
2cosx-1=o
cosx=1/2

so.. x=o, pi, pi/3, & 5pi/3

is that right? thank you for your help!

 

[Unco]

Nicely done.