Solve on the interval [0,2pi)
(sinx-1)(2sin^2 x-5sinx+2)
I know a little bit about how to solve on the interval, but everything that I have tried is not getting me an answer-not sure what else I could try... Thanks!

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you forgot to put the rest of the equation

The equation is on the second line, but here it is again: (sinx-1)(2sin^2 x-5sinx+2)

but it has to equal something to solve it

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