Question

solve sinx+sin2x=0 for [0,2pi)

Answer 1

What's there to solve? X=0,pi,2pi? Is that what you want?

Answer 2

all the answers between [0,2pi)

Answer 3

sin2x=2sinxcosx
sinx+sin2x=sinx+2sinxcosx= sinx(1+2cosx)=0

Answer 4

now we need to solve
sinx=0 and 1+2cosx=0

Answer 5

when sinx=0 on a unit circle we get x=0 and x=pi

Answer 6

when cosx=-1/2 we get x=2pi/3 and x=4pi/3

Answer 7

so you get four answers total, hope you get it;)

Answer 8

thanks