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Problem in finding the General Solution of a Trigonometric Equation v2

  1. May 19, 2017 #1
    1. The problem statement, all variables and given/known data:

    Find the general solution of the equation: $$\tan {x}+\tan {2x}+\tan {3x}=0$$

    Answer given: ##x=## ##\frac {n\pi}{3}##, ##n\pi \pm \alpha## where ##\tan {\alpha} = \frac {1}{\sqrt {2}}##.

    2. Relevant equations:

    These equations may be used:

    20170519_023122.png

    3. The attempt at a solution:

    Please see the pic below:

    1495188863110-223051699.jpg

    The answer from the "EITHER" is correct, but how do I simplify the second part?
     
    Last edited: May 19, 2017
  2. jcsd
  3. May 19, 2017 #2

    BvU

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    I find it hard to read and it is badly focused, so I am reluctant to type it out (after all you are too lazy to do so too ...), but I can distinguish you go from $$\left [ 1\ + \ {1\over 1-\tan\alpha\tan 2\alpha} \right ] = 0 $$ to something I don't understand and -- If I read it right -- don't believe...

    Could you post the steps (typed) ?
     
  4. May 19, 2017 #3
    I've posted a better picture. Please see. And the ##\alpha## in your post will be ##x##.
     
  5. May 19, 2017 #4

    BvU

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    Better focused, yes. Understand or believe ?
     
  6. May 19, 2017 #5

    haruspex

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    Looks good as far as you went, but why did you stop?
    You can throw away the denominator, and convert the cos 3x back to trig terms in x and 2x again.
     
  7. May 19, 2017 #6

    BvU

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    Ah, the last ##\cos 3x## is more like ##\cos 3x## plus .....
     
  8. May 19, 2017 #7
    I could, but I would've also got a sin term, which I can take to other side, and then cross multiply to get tan terms in x and 2x. Then?
     
  9. May 19, 2017 #8

    haruspex

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    You should get an equation involving tan x, tan 2x and a constant.
     
  10. May 19, 2017 #9
    Yes, ##2\tan {x}\tan {2x}=0##.

    Then....I understood. Thanks a lot.
     
  11. May 19, 2017 #10

    haruspex

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    did you mean, ##2-\tan {x}\tan {2x}=0##?
     
  12. May 19, 2017 #11
    No, I did a wrong calculation. Then how will I proceed after that?
     
  13. May 19, 2017 #12

    BvU

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    I second haru: ##2-\tan {x}\tan {2x}=0## to be solved . Repeat: see post #2.
     
  14. May 19, 2017 #13

    haruspex

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    If you now have that equation, expand the tan 2x. If not, please post your working.
     
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