Question

solve on the interval [0,2pi)
(sinx-1)(2sin^2x-5sinx+2)=0

Answer 1

You have two separate equations to look at:
sinx - 1 = 0
and
2sin^2x - 5sinx + 2 = 0

Answer 2

Because if you can make one of those quantities equal 0 with some x, you make the whole thing 0

Answer 3

so set both equal to zero?

Answer 4

Yes, and find the solutions for each

Answer 5

The first one is easier so I'll give it to you. x needs to be pi/2