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

x = π / 6 ,  π / 2  and 5 π / 6. 

I’d say d hope it