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Homework Help: Problem in finding a general solution

  1. Jul 13, 2018 #1


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    1. The problem statement, all variables and given/known data
    thumbnail_IMG_20180713_225234 (1).jpg

    2. Relevant equations
    General Formula for Tan(a)=Tan(b)

    3. The attempt at a solution
    See the question I have uploaded.

    I have tried solving it this way,

    Firstly I applied the Quadratic Formula to get,


    Now we have two cases,


    When 12%29%3D2-%5Csqrt3.gif

    So General Formula here will be,


    Now, CASE-2

    when 12%29%3D-%282+%5Csqrt3%29.gif

    So General Formula here will be 12.gif

    I do not know what should I do next to get the answer? Please tell me how to proceed Further.

    The answer given in the key is the option (C).

    I will be thankful for any help!
  2. jcsd
  3. Jul 13, 2018 #2


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    2017 Award


    So you need to combine ##\pi (n+ {1\over 12}) ## and ##\pi (n - {5\over 12}) ## . Leave the ##\pi## outside the brackets and try a few n. The pattern emerges !
  4. Jul 13, 2018 #3


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    if the 6n looks scary in the option C, it is actually just ##n/2## cause ##a=(6n+1)\frac{\pi}{12}=\frac{n}{2}\pi+\frac{\pi}{12}##. From this very last expression for a, what do you get if you put
    1) n=even=2k
    2) n=odd=2k+1
  5. Jul 14, 2018 #4


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  6. Jul 15, 2018 #5


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    I think the problem is solved by @Delta2's hint, so I may come with a very simple solution of the equation tan2(α)+2√3 tan(α)=1,which can be rearranged to 1-tan2(α)=2√3 tan(α).
    nπ/2 in the offered solutions suggests to solve the equation for 2α. The double-angle formula is ##\tan(2α)=\frac{2\tan(α)}{1-\tan^2(α)}##, that is ##\tan(2α)=\frac{2\tan(α)}{2\sqrt3 \tan(α)}=\frac{1}{\sqrt 3}##, that is, 2α=π/6+kπ and α=π/12+kπ/2.
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