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Trigonometric Equation

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data
    cos(x)+2tg(x) = 7/(4*cosx(x)) Solve for x on the real numbers set.


    2. Relevant equations
    tg(x) = sin(x)/cos(x) ,
    cos^2(x) = 1-2*sin^2(x) ,
    The determinant is D = b^2 - 4ac ,
    Also cos(x) ,tg(x), sin(x) | -1 < x < 1


    3. The attempt at a solution
    For final arrangement I got is:
    -8sin^2(x) + 8sin(x) - 3 = 0

    But the determinant is 8^2 - 4*(-8)*(-3) which is 64 - 96 thus the equation doesn't have a real solution.

    I checked on WolframAlpha and it has some weird solutions and I was also told that it does have a real solution.
     
  2. jcsd
  3. Apr 7, 2010 #2

    ehild

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    The coefficient 8 in front of sin^2(x) is wrong.

    ehild
     
  4. Apr 7, 2010 #3
    Why?
    Let's say we rearrange the original equation to: cos^2(x) + 2 sin(x) = 7/4
    // We multiplied by cosx(x) //
    Then cos^2(x) becomes 1-2sin^2(x)... That multiplied by 4 will be 4-8sin^2(x).
    I don't think where I got that wrong.
     
  5. Apr 7, 2010 #4

    tiny-tim

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    Hi Icelove! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    nooo :redface: … learn your trigonometric identities …

    cos2x = 1 - 2sin2x :wink:
     
  6. Apr 7, 2010 #5

    ehild

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    [tex]\cos{x}\cdot\cos{x}=(\cos{x})^2=1-(\sin{x})^2[/tex]

    ehild
     
  7. Apr 7, 2010 #6
    cos(x)+2tg(x) = 7/(4*cosx(x)) Solve for x on the real numbers set.

    By this you the complete equation is

    [tex](4 \cdot cos(x)^2)\cdot cos(x) + (4\cdot cos(x)^2)\cdot (2tan(x)) - 7 = 0 [/tex] which implies that

    [tex] (4 \cdot cos(x)^2) \cdot (cos(x) + 2 tan(x)) - 7 = 0[/tex]
     
    Last edited: Apr 7, 2010
  8. Apr 7, 2010 #7
    ooooooooooooooooohh... Thanks. :D
     
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