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Basic Trig equation

  1. Feb 23, 2006 #1

    Hootenanny

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    The curve [itex] y = \cos 3x + 2[/itex] intersects the line [itex]y = 2x[/itex] at point [itex]A[/itex], whose x co-ordinate is [itex]\alpha[/itex]. Show that [itex] 0.7 < \alpha < 0.8 [/itex].

    So far I've got: Upon intersection [itex]2x = \cos 3x + 2 \Rightarrow \cos 3x - 2x = - 2 [/itex]. This doesn't seem to help. I know we've done this type of thing ages ago, but I've since lost my notes and my minds gone blank. Any help would be appreciated.
     
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  3. Feb 23, 2006 #2

    Galileo

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    How about evaluating both functions at x=0.7 and x=0.8? See what you can do with that.
     
  4. Feb 23, 2006 #3

    Hootenanny

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    [itex]\cos(3 \times 0.7) +2 = 1.495...[/itex] , [itex]\cos(3 \times 0.8) +2 = 1.262...[/itex].
    [itex]2 \times 0.7 = 1.4[/itex], [itex]2 \times 0.8 = 1.6[/itex].
    All in radians. This doesnt seem to help??
     
  5. Feb 23, 2006 #4

    arildno

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    Sure it does!
    At x=0.7, we have the value as given by the straight line LOWER than that given by the cosine expression, whereas this is reversed at x=0.8
    What does that tell you?
     
  6. Feb 23, 2006 #5

    Hootenanny

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    Ahhh, ofcourse! Tha x - value must lie sumwhere between them values! I wan looking for an exact solution. Thank's foryou help guys!
     
  7. Feb 23, 2006 #6

    arildno

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    You're welcome.
    Most equations cannot be solved for an exact solution in a finite number of steps.
    Approximative techniques abound, though.
     
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