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jkl3inberg Group Title

Find all solutions of the equation in the interval [0,2pi). Write your answer in terms of pi. √3 cot(theta)-1=0 I got cot(theta)= pi/3, but the identity of cotangent is x/y, so I'm not sure where to look at. Is the answer just pi/3?

  • one year ago
  • one year ago

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  1. hartnn Group Title
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    did you mean you got thetha = pi/3 ?

    • one year ago
  2. NoelGreco Group Title
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    That's one answer, but there is another angle on that interval where the cotangent is positive and the reference angle is also pi/3.

    • one year ago
  3. jkl3inberg Group Title
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    Yes hartnn, and NoelGreco the only place there's a reference angle of pi/3 is 1/2 and root3/2, are those my answers?

    • one year ago
  4. hartnn Group Title
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    he meant there are 2 solutions you got cot theta =1/ sqrt 3 right ? there are 2 angles in [0,2pi) which satisfies this, one angle is pi/3 which you got it correct and other will be (pi/3 +pi) =... ? because cot function is periodic in pi

    • one year ago
  5. jkl3inberg Group Title
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    I got 1/sqrt 3 yes, but that needs to be rationalized to root 3 / 3, right? Which I don't see anywhere on the unit circle

    • one year ago
  6. hartnn Group Title
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    oh, you are using unit circle, ok then cot = cos / sin search for the values where the ratio of cos/ sin would equal 1/sqrt 3 one when cos = 1/2 and sin = sqrt 3 /2 and one more when cos =-1/2 and sin = -sqrt 3/2 those 2 angles will be your solution

    • one year ago
  7. jkl3inberg Group Title
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    so pi/3 and 4pi/3, thanks. I get these kinds of problems but cot=x/y really confused me. How exactly did you end up with 1/2 and root3/2?

    • one year ago
  8. hartnn Group Title
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    ok, cot (theta) = 1/ sqrt 3 is the ratio of cos and sine functions so, for sin and cos the standard values are 1/2 and sqrt 3/2 for the standard angle pi/3, (we need to remember standard angles and their values or we can use the unit circle) ratio of only these would give us 1/sqrt 3...thats how.

    • one year ago
  9. jkl3inberg Group Title
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    Ok, got it. Thanks!

    • one year ago
  10. hartnn Group Title
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    welcome ^_^

    • one year ago
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