Question

How do you solve the trig equation: `sin^2x + sinx = 0` from [0,2pi]?

Answer 1

You can write it, collecting `sin(x)`, as:
`sin(x)[sin(x)+1]=0`
So:
`sin(x)=0` for `x=0, pi, 2pi`
and
`sin(x)=-1` for `x=3/2pi`