Solve on the interval [0,2pi)

(sinx-1) (2sin^2 x-5sinx+2)

Sin x - 1 = 0

sin x = 1

x 1 = π / 2

2 sin² x + 5 sin x + 2 = 0

Substitution: u = sin x

2 u² - 5 u + 2 = 0

2 u² - 4 u - u + 2 = 0

2 u (u - 2) - (u - 2) = 0

(u - 2) (2 u - 1) = 0

(sin x - 2) (2 sin x - 1) = 0

sin x - 2 = 0

sin x = 2 (does not have solution)

2 sin x - 1 = 0

2 sin x = 1

sin x = 1/2

x 2 = π / 6, x 3 = 5 π / 6

Answer:

x = π / 6, π / 2 and 5 π / 6.