Find all solutions in the interval [0, 2π). sin²x + sin x = 0

Solution:

It is given that

sin²x + sin x = 0, [0, 2π)

Let us take sin x as common

sin x(sin x + 1) = 0

sin x = 0

So we get

x = 0, (π - 0), (π + 0), (2π - 0) 

or 

sin x + 1 = 0

sin x = -1

x = π + π/2

x = 0, 3π/2, π

Therefore, all the solutions are x = 0, 3π/2, π.

Find all solutions in the interval [0, 2π). sin²x + sin x = 0

Summary:

All solutions in the interval [0, 2π). sin²x + sin x = 0 are x = 0, 3π/2, π.