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Solving a trigonometric equation

  1. Jan 13, 2009 #1
    1. The problem statement, all variables and given/known data

    sqrt3 cot 3x + 1 = 0 where 0 _<x_<2pie


    2. Relevant equations



    3. The attempt at a solution

    I narrowed it down to a constant on one side,

    cot 3x = -1/sqrt3 which I could turn into cos x/sin x = -1/sqrt3

    but I have no idea where to go from here.

    Any help would be greatly appreciated!
     
  2. jcsd
  3. Jan 13, 2009 #2

    gabbagabbahey

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    Hmmm... if [tex]\cot(3x)=\frac{-1}{\sqrt{3}}[/tex], then [tex]\tan(3x)=???[/tex]:wink:
     
  4. Jan 13, 2009 #3
    well it would be 1/-1/sqrt3 wouldnt it?
     
  5. Jan 13, 2009 #4

    gabbagabbahey

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    Sure, but [tex]\frac{1}{\frac{-1}{\sqrt{3}}}=-\sqrt{3}[/tex] right? :wink:

    Do you know what angle(s) has a tangent of [itex]-\sqrt{3}[/itex]?
     
  6. Jan 13, 2009 #5
    I dont actually. Im only familiar with the unit circle for sin/cos.
     
  7. Jan 13, 2009 #6

    jgens

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    As tan(x) is defined as sin(x)/cos(x) you should be able to find the values for tan(x).
     
  8. Jan 13, 2009 #7

    gabbagabbahey

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    Try checking every 30 degrees ([itex]\pi/6[/itex] radians) between 0 and 180 degrees. You should find one value in that range.

    Then you know that you want to find all values for [itex]0\leq x\leq2\pi[/itex], so that means that 3x goes from 0 to 6pi Right?

    That means you need to find all angles between zero and 6pi where the tangent of those angles is -sqrt{3}. You can use the fact that Tangent is periodic to do that by adding multiples of 180 degrees o the angle you found earlier.
     
  9. Jan 13, 2009 #8
    okay well one second here, to find values of tan x using sin x/cos x, do i calculate that with a calculator to find the angle or use the unit circle, cause im looking at the unit circle and there are no values for sin or cos sqrt3.
     
  10. Jan 13, 2009 #9

    gabbagabbahey

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    use the unit circle: there are values where [tex]\sin(\theta)=\frac{\sqrt{3}}{2}[/tex] and [tex]\cos(\theta)=\frac{-1}{2}[/tex] aren't there? :wink:

    Follow my instructions from my previous post; what is [itex]\tan(0)[/itex]? How about [itex]\tan(\pi/6)[/itex]? How about [itex]\tan(\pi/3)[/itex] ....etc.?
     
  11. Jan 13, 2009 #10
    I dont get how you get those sin and cos values from tan sqrt 3.

    Like using a calculator I get a value of -60. Then adding 180 degrees I get -60, 120, 300, 480, 660, 840, and 1020 for 3x.
     
  12. Jan 13, 2009 #11

    gabbagabbahey

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    Just plug in angles every 30 degrees from 0 to 180....you'll see where i get them from:wink:
     
  13. Jan 13, 2009 #12
    Whats the process of plugging in angles every 30 degrees for tan 3x = sqrt3?
     
  14. Jan 13, 2009 #13

    gabbagabbahey

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    [tex]\tan(0)=\frac{\sin(0)}{\cos(0)}=\frac{0}{1}=0[/tex]

    [tex]\tan(30)=\frac{\sin(30)}{\cos(30)}=\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}[/tex]

    [tex]\tan(60)=???[/tex]

    [tex]\tan(90)=???[/tex]

    [tex]\tan(120)=???[/tex]

    [tex]\tan(150)=???[/tex]

    [tex]\tan(180)=???[/tex]
     
  15. Jan 13, 2009 #14
    Okay I gotcha there.

    So the angles ive found are correct then? 120, 300? I assume I dont include the -60 because our range is 0 to 2pie?
     
  16. Jan 13, 2009 #15

    gabbagabbahey

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    If 3x=120, then x=40...you should end up with 6 values for x in between 0 and 2pi
     
  17. Jan 13, 2009 #16
    So with this then, I have my two angles of 2pie/3 and 5pie/3, is 3x 2pie/6 5pie/6 and 2pie/9 and 5pie/9? Are those my 6 angles?
     
  18. Jan 13, 2009 #17
    OK just have to say this, pi not pie
     
  19. Jan 13, 2009 #18
    ahhaha, im sorry i should know better.
     
  20. Jan 14, 2009 #19
    Is my answer is post #16 correct? Or on the right track at least?
     
  21. Jan 14, 2009 #20
    The beauty of math at your level is that you can just plug your answer in and see if it's right.
     
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