Question

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!

Answer 1

you forgot to put the rest of the equation

Answer 2

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

Answer 3

but it has to equal something to solve it

Answer 4

Hmm...It's not set to equal anything... but maybe equals x?

Answer 5

@dpalnc...If i factor it will I have to solve more after that in order to get to the answer?

Answer 6

@missfitz5172 , forget that... I misread "interval" as "integrate".... hehe :)

Answer 7

do you want the zeros of this graph? i.e., where the graph crosses the x-axis?

Answer 8

@dpalnc, I'm looking for the solutions. I think that if this were graphed, that the zeroes would be the same as the solutions, so I think so...

Answer 9

the problem does not state an equation to solve. is this the equation:
(sinx-1)(2sin^2 x-5sinx+2) = 0

Answer 10

@dpaInc, yes that is the equation

Answer 11

if it is, then factoring will definitely help.

Answer 12

factor this part of the expression: (2sin^2 x-5sinx+2)

Answer 13

@dpaInc , (2sin^2 x-5sinx+2) factored to (2sinx-1)(sinx-2)

so does that mean that all together I have (sinx-1)(2sinx-1)(sinx-2) ?

Answer 14

yep... good

Answer 15

notice that last factor (sinx -2) will not help any...

Answer 16

ok, now my answer has to be in a form on the unit circle...how can i translate these factored groups to something in a fraction with pi?

Answer 17

can you explain why the (sinx-2) doesn't help?

Answer 18

you'll need to set each factor to 0:
sinx - 1 = 0 and 2sinx - 1 = 0
let's work on the first one...