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Trigonometry inequalities

  1. Apr 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the values for x.

    [tex]3sin^2x - 3sinxcosx + 2cos^2x > 1[/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex]3sin^2x - 3sinxcosx + 2cos^2x > 1[/tex]

    [tex]3sin^2x - 3sinxcosx + 2cos^2x > sin^2x+cos^2x[/tex]

    [tex]2sin^2x - 3sinxcosx + cos^2x > 0[/tex]

    What to do next?
     
  2. jcsd
  3. Apr 26, 2008 #2
    You can use the same formula sin^2+cos^2=1 again. Now rewrite the equation in terms of only one trigonometric function.
     
  4. Apr 26, 2008 #3

    Tom Mattson

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    I think it would be easier to just factor it the way it is.
     
  5. Apr 26, 2008 #4
    What to do now?
    If I try with [itex]sin^2+cos^2=1[/tex]

    I would get:
    [tex]sin^2x-3sinxcosx>-1[/tex]

    Tom Mattson:
    [tex](sinx-cosx)(2sinx-cosx)>0[/tex]
    I am wondering why in my book says, I am giving quote:

    Hm...
     
  6. Apr 26, 2008 #5

    Tom Mattson

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    Your book probably took a different approach. Doing it my way, the two cases are:

    1.) [itex]\sin(x)-\cos(x)>0[/itex] and [itex]2\sin(x)-\cos(x)>0[/itex], or

    2.) [itex]\sin(x)-\cos(x)<0[/itex] and [itex]2\sin(x)-\cos(x)<0[/itex].
     
  7. Apr 26, 2008 #6
    But they have 3 solutions for x.
    The first: [tex]x=\frac{\pi}{2}+k\pi[/tex]
    The second: [tex]tgx<\frac{1}{2}[/tex]
    The third: [tex]tgx>1[/tex]

    In your case we have only 2:
    [tex]tgx<\frac{1}{2}[/tex]
    [tex]tgx>1[/tex]
     
  8. Apr 26, 2008 #7

    Tom Mattson

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    No, Physicsissuef. When you rearranged the inequalities, you divided by [itex]\cos(x)[/itex]. That means you obtained results that are valid only for [itex]\cos(x)\neq 0[/itex]. To find out what happens when [itex]\cos(x)=0[/itex], you have to examine those points separately.

    So, for which [itex]x[/itex] is [itex]\cos(x)=0[/itex], and do the inequalities hold for all such [itex]x[/itex]?
     
  9. Apr 26, 2008 #8
    But both points for [itex]cosx=0[/itex] and [itex]\cos(x)\neq 0[/itex] are valid.
    For pi/2 and -pi/2 cosx=0
     
  10. Apr 26, 2008 #9

    Tom Mattson

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    That's exactly what I just told you. When you divided by [itex]\cos(x)[/itex], you obtained results that are valid only for [itex]\cos(x)\neq 0[/itex]. You then have to go back and find out if the inequalities hold for all [itex]x[/itex] such that [itex]\cos(x)=0[/itex].
     
  11. Apr 26, 2008 #10
    but why not sin(x)=0, that's my question, or why not sin(x)=1 or cos(x)=1, etc.etc, how will I know which of all of this values to choose?
     
  12. Apr 26, 2008 #11

    Tom Mattson

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    You didn't divide by [itex]\sin(x)[/itex], did you?

    Question: What did you divide by?

    Answer: [itex]\cos(x)[/itex].

    Question: So what are the only values of [itex]x[/itex] for which that division is not defined?

    Answer: The values of [itex]x[/itex] for which [itex]\cos(x)[/itex], the thing you divided by, is zero.
     
  13. Apr 26, 2008 #12
    what means "the division is not defined"?
    Do you mean [itex]\frac{sinx}{cosx}=tgx[/itex], so for cosx=0, there will be infinity, but if it is only cosx=0, there will be no infinity right?
     
  14. Apr 27, 2008 #13

    tiny-tim

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    Hi Physicsissuef! :smile:

    Solutions 2 and 3 are in terms of tanx, and technically tanx is undefined at multiples of π/2.

    For example, just below π/2, it's huge and positive, but just above π/2 it's huge and negative. So at π/2 it's both-infinity-and-minus-infinity … which doesn't help with tanx > 1, does it? :frown:

    So they had to add Solution 1, just to cover all possibliities! :smile:
     
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