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Math Help - Trigonometry equation 2

  1. #1
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    Trigonometry equation 2

    4cos^23x=1
    cos^23x=1/4
    (cos^22\pi)+cos^2x=1/4 can i do a square root here?
    assuming I cant I went on and did
    (2cos^2x-1)^2+cos^2x=1/4
    [tex](4cos^4x-2cos^2x+1)+cos^2x=1/4[\math]
    4cos^4x-cos^2x+1=1/4
    cos^2x(4cos^2x-1)+1=1/4

    how .. do I do this.. especially without the unit circle
    I looked at the answers and the values are not on the unit circle.
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  2. #2
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by dorkymichelle View Post
    4cos^23x=1
    cos^23x=1/4
    (cos^22\pi)+cos^2x=1/4 can i do a square root here?
    assuming I cant I went on and did
    (2cos^2x-1)^2+cos^2x=1/4
    [tex](4cos^4x-2cos^2x+1)+cos^2x=1/4[\math]
    4cos^4x-cos^2x+1=1/4
    cos^2x(4cos^2x-1)+1=1/4

    how .. do I do this.. especially without the unit circle
    I looked at the answers and the values are not on the unit circle.
    Here is how I would approach the problem

    4\cos^2{3x}=1

    \cos^2{3x}=\frac{1}{4}

    now take the square root

    \cos{3x} = -\frac{1}{2} or \cos{3x} = \frac{1}{2}

    Can you finish up from here?
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  3. #3
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    The next step I got is
    cos2x+cosx=1/2
    [tex]2cos^2x-1+cosx=1/2
    2cos^2x+cosx-3/2=0
    cosx(2cosx+1)-2/3=0
    cosx = 1/2
    so \pi/3 + n\pi
    but the book's answer is  \pi/9+2\pi n/3, 5\pi/9+2\pi n/3,2\pi/9+2\pi n/3, 4\pi/9+2\pi n/3

    How do I get all these answers?!
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  4. #4
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    \cos{(3x)} does NOT equal \cos{x} + \cos{(2x)}.

    So far you have two possibilities:

    Case 1:

    \cos{(3x)} = \frac{1}{2}.

    Cosine is positive in the first and fourth quadrants, and to get to x, you have to "undo" the cosine - so take the arccos (or \cos^{-1}...


    3x = \arccos{\frac{1}{2}}

    3x = \left \{ \frac{\pi}{3}, 2\pi - \frac{\pi}{3} \right \} + 2\pi n, n \in \mathbf{Z}, since the period of cosine is 2\pi.

    3x = \left \{ \frac{\pi}{3}, \frac{5\pi}{3} \right \} + 2\pi n

    x = \left \{ \frac{\pi}{9}, \frac{5\pi}{9} \right \} + \frac{2\pi n}{3}.


    Case 2:

    \cos{(3x)} = -\frac{1}{2}

    3x = \arccos{\left(-\frac{1}{2}\right)}

    3x = \left \{ \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3} \right \} + 2\pi n, since cosine is negative in the second and third quadrants and the period is 2\pi


    3x = \left \{ \frac{2\pi}{3}, \frac{4\pi}{3} \right \} + 2\pi n

    x = \left \{ \frac{2\pi}{9}, \frac{4\pi}{9} \right \} + \frac{2\pi n}{3}.


    So putting it all together, the possibilities are

     \left \{ \frac{\pi}{9}, \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{5\pi}{9} \right \} + \frac{2\pi n}{3}.
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