anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
you forgot to put the rest of the equation
anonymous
  • anonymous
The equation is on the second line, but here it is again: (sinx-1)(2sin^2 x-5sinx+2)
anonymous
  • anonymous
but it has to equal something to solve it

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anonymous
  • anonymous
Hmm...It's not set to equal anything... but maybe equals x?
anonymous
  • anonymous
@dpalnc...If i factor it will I have to solve more after that in order to get to the answer?
anonymous
  • anonymous
@missfitz5172 , forget that... I misread "interval" as "integrate".... hehe :)
anonymous
  • anonymous
do you want the zeros of this graph? i.e., where the graph crosses the x-axis?
anonymous
  • anonymous
@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...
anonymous
  • anonymous
the problem does not state an equation to solve. is this the equation: (sinx-1)(2sin^2 x-5sinx+2) = 0
anonymous
  • anonymous
@dpaInc, yes that is the equation
anonymous
  • anonymous
if it is, then factoring will definitely help.
anonymous
  • anonymous
factor this part of the expression: (2sin^2 x-5sinx+2)
anonymous
  • anonymous
@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) ?
anonymous
  • anonymous
yep... good
anonymous
  • anonymous
notice that last factor (sinx -2) will not help any...
anonymous
  • anonymous
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?
anonymous
  • anonymous
can you explain why the (sinx-2) doesn't help?
anonymous
  • anonymous
you'll need to set each factor to 0: sinx - 1 = 0 and 2sinx - 1 = 0 let's work on the first one...|dw:1333235792196:dw|
anonymous
  • anonymous
ok, lemme explain about that last factor, sinx - 2. when you solve sinx -2 = 0, you'll get: sinx = 2. look at the unit circle. what angle can you take the sine of to get 2?
anonymous
  • anonymous
sinx-1=0 equals x=pi/2 Could it be either pi/6, pi/4, or pi/3? (about what sine to get 2)
anonymous
  • anonymous
notice that sine can never be bigger than 1 or smaller than -1. that's why sinx = 2 does not have a solution.
anonymous
  • anonymous
ooohhh...i apologize, i'm not good at math haha. Ok, so sinx-1=0 equals x=pi/2
anonymous
  • anonymous
that's ok... i'm not good either!!!
anonymous
  • anonymous
but the answer about that first factor is yes, x=pi/2 now how about that second factor...
anonymous
  • anonymous
2sinx - 1 = 0 factors to be x = pi/6
anonymous
  • anonymous
is that it? there's 1 more you forgot..
anonymous
  • anonymous
the (sin x -2)?
anonymous
  • anonymous
|dw:1333236380409:dw|
anonymous
  • anonymous
Ok... so i'm not sure which factor that is, but is it 5pi/6?
anonymous
  • anonymous
that should be it then since sinx - 2 doesnot have a solution...
anonymous
  • anonymous
ok, so then my answer should be pi/2, pi/6, and 5pi/6...
anonymous
  • anonymous
yep, in the interval 0 to 2pi....
anonymous
  • anonymous
wow. thank you so much for your help!! really really, it means a lot (: thanks again (:
anonymous
  • anonymous
way to stick with it... and good job. :)
anonymous
  • anonymous
i really appreciate that you explained everything too (: and thanks!

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