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Trigonometric equation

  1. Jun 2, 2004 #1
    How to solve this equation:
    tan 2x + sec 2x = cos x + sin x where 0<=x<=360
    I solve it in this way but cannot find the right answer:
    (sin 2x + 1) / cos 2x = cos x + sin x
    (cos x + sin x)^2 / cos 2x = cos x + sin x
    (cos x + sin x) / cos 2x = 1
    sqrt(2) cos(x - 45) sec 2x = 1
    sec 2x cos(x - 45) = 1 / sqrt(2)

    x = 90 or 360

    But the correct answers are x = 0 or 270 or 360.
     
  2. jcsd
  3. Jun 2, 2004 #2
    (cos x + sin x)2 equals one, how did you figure that it equals (sin 2x + 1)?
     
  4. Jun 2, 2004 #3

    Hurkyl

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    Gold Member

    That seems a pretty big leap; maybe if you finished doing the work on it?



    You're thinking of cos2x + sin 2 x.
     
  5. Jun 2, 2004 #4
    :rolleyes: Thanks.
     
  6. Jun 2, 2004 #5
    I have solved the problem, I have made some careless mistakes above.
    The solution is:
    tan 2x + sec 2x = cos x + sin x where 0<=x<=360
    (sin 2x + 1) / cos 2x = cos x + sin x
    (cos x + sin x)^2 / cos 2x = cos x + sin x
    (cos x + sin x) / cos 2x = 1
    (cos x + sin x) / [(cos x + sin x)(cos x - sin x)] = 1
    cos x - sin x = 1
    sqrt(2) cos(x + 45) = 1
    cos(x + 45) = 1 / sqrt(2)
    Then x = 0 or 270 or 360
     
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