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Thread: Trigonometry equation 2

  1. #1
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    Trigonometry equation 2

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

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

    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

    $\displaystyle 4\cos^2{3x}=1 $

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

    now take the square root

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

    Can you finish up from here?
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  3. #3
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    The next step I got is
    $\displaystyle cos2x+cosx=1/2$
    [tex]2cos^2x-1+cosx=1/2
    $\displaystyle 2cos^2x+cosx-3/2=0$
    $\displaystyle cosx(2cosx+1)-2/3=0$
    $\displaystyle cosx = 1/2$
    $\displaystyle so \pi/3 + n\pi$
    but the book's answer is $\displaystyle \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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    $\displaystyle \cos{(3x)}$ does NOT equal $\displaystyle \cos{x} + \cos{(2x)}$.

    So far you have two possibilities:

    Case 1:

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

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


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

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

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

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


    Case 2:

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

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

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


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

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


    So putting it all together, the possibilities are

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