Answer:
x = 0, π, 3π/2.
Explanation:
We can factor the left-hand side of the equation:
sin²x + sin x = sin x (sin x + 1) = 0
Therefore, either
sin x = 0 or sin x + 1 = 0.
For sin x = 0, the solutions over the interval [0, 2π) are
x = 0, π.
For sin x + 1 = 0,
we have sin x = -1.
This only occurs at x = 3π/2.
Therefore, the exact solutions of the equation sin²x + sin x = 0 over the interval [0, 2π) are:
x = 0, π, 3π/2.