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Yahoo!

Find the general solution of tan2x = -cot(x+pie/3)?

  • one year ago
  • one year ago

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  1. A.Avinash_Goutham
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    u kno tan2x = 2tanx/1-tan^2x nd cotx = 1/tanx

    • one year ago
  2. A.Avinash_Goutham
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    form a quadratic nd solve........ for tanx........ then u l get the g.s

    • one year ago
  3. waterineyes
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    \[\tan(2x) = -\cot(90 + x -30) \implies -(-\tan(x-30)) = \tan(x - \frac{\pi}{6})\] \[\implies \tan(2x) = \tan(x - \frac{\pi}{3})\]

    • one year ago
  4. waterineyes
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    General solution is given by: \[\tan ( \theta) = \tan(x)\] then: \[\theta = n \pi + x\]

    • one year ago
  5. waterineyes
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    so: \[2x = n \pi + ( x - \frac{\pi}{3} ) \implies x = n \pi + \frac{\pi}{3}\]

    • one year ago
  6. Yahoo!
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    hey @waterineyes but my books says it as x = n pie + 5pie/6

    • one year ago
  7. waterineyes
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    So put n = 0 1 2 3 4 etc etc and find the corresponding value of x..

    • one year ago
  8. waterineyes
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    you can subtract it from \(2 \pi\)..

    • one year ago
  9. Yahoo!
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    wat???

    • one year ago
  10. waterineyes
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    Wait something got wrong.. Let me check..

    • one year ago
  11. Yahoo!
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    @w\ r uthere

    • one year ago
  12. waterineyes
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    Wait I am checking it..

    • one year ago
  13. waterineyes
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    I got it..

    • one year ago
  14. sauravshakya
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    |dw:1343210793286:dw|

    • one year ago
  15. waterineyes
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    \[\tan(2x) = \cot(-x-60) \implies \cot(90 - (x + 150)) \implies \tan(x + 150)\] \[\tan(2x) = \tan(x + 150)\]

    • one year ago
  16. waterineyes
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    150 is in degrees: \[150^{\circ} = \frac{5 \pi}{6}\]

    • one year ago
  17. waterineyes
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    So: \[2x = n \pi + x + \frac{5 \pi}{6}\]

    • one year ago
  18. Yahoo!
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    wat the wrong in 1 st method!!

    • one year ago
  19. waterineyes
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    and subtract x from both sides: \[x = n \pi + \frac{5 \pi}{6}\]

    • one year ago
  20. waterineyes
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    My mistake in reading i read as : x - pie/3

    • one year ago
  21. Yahoo!
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    thxxx

    • one year ago
  22. waterineyes
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    Welcome dear..

    • one year ago
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