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Math Help - why equation, wrong

  1. #1
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    why equation, wrong

    4 sin^2(x) + 8 cos(x) = 7

    1 = cosē(x)+sinēx
    sinē(x) = 1-cosē(x)

    we use this and replace sinē(x) with 1-cosē(x)

    4(1-cosē(x))+8cos(x) = 4-4cosē(x)+8cos(x)

    -4cosē(x)+8cos(x)+4 = 7

    replace
    [cos(x)=t] and multiplicate with -1

    4tē-8t-4 = -7 <--> t1= 1/2 t2=3/2

    t = cos(x),

    cos(x1)=1/2
    cos(x2)=3/2

    x1=pi/3 + n *2pi or x1=pi - pi/3 + n *2pi but the second one is wrong ? why ?
    Cos(x2) > 1 = dont work
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  2. #2
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    e^(i*pi)'s Avatar
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    You're working is fine - it's a domain and range issue - you can discard t2

    The maxiumum value of \cos(x) is 1. Since \dfrac{3}{2} > 1 then you cannot take the inverse cos.


    It's not in the range of f(x) = \cos(x) and because cos(x) and arccos(x) are inverses of each other then the range of f(x) becomes the domain of f^{-1}(x) which means it's not in the domain of f^{-1}(x)
    Last edited by e^(i*pi); January 15th 2011 at 01:33 PM. Reason: explaining domain/range and inverse functions
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  3. #3
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    yes i know that i meant that the wrong is in x1=pi - pi/3 + n *2pi
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  4. #4
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    i think maybe it will be x1=pi/3 + n *2pi or x1=pi - pi/3 - n *2pi

    right now ?
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  5. #5
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    Quote Originally Posted by paulaa View Post
    4 sin^2(x) + 8 cos(x) = 7
    ...
    cos(x1)=1/2

    x1=pi/3 + n *2pi or x1=pi - pi/3 + n *2pi but the second one is wrong ? why ?
    cos(x) is an even function. If cos(x1) = 1/2, then also cos(-x1) = 1/2

    Therefore the second one should be: -x1 = pi/3 → x1 = -p1/3. Of course, you can add n*2pi
    So the final form for the second solution is: x1 = pi/3 + n 2pi.
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