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Identifying x-values

  1. Aug 18, 2005 #1
    I am trying to identify the x-values that are solutions for the equation:
    8cosx-4=0

    I have come up with: x=pi/3,4pi/3
    This doesn't seem correct but I am stumped. :confused:
     
  2. jcsd
  3. Aug 18, 2005 #2

    VietDao29

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    Homework Helper

    Why 4pi/3?
    Remember, if you need to solve:
    [tex]\cos \alpha = \cos \beta = x[/tex]
    [tex]\Leftrightarrow \alpha = \pm \beta + k2\pi, k \in \mathbb{Z}[/tex]
    In other words,
    [tex]\alpha = \pm \arccos x + k2\pi, k \in \mathbb{Z}[/tex]
    Viet Dao,
     
  4. Aug 18, 2005 #3
    so,
    5pi/3

    I C

    Thank you
     
  5. Aug 18, 2005 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    8 cos x= 4 so cos x= 1/2. You could use a calculator but I think of half an equilateral triangle to observe that [itex]cos\left(\frac{\pi}{3}\right)= \frac{1}{2}[/itex]. I then recall that "cos t" is the x coordinate of the unit circle parametrized by x= cos t, y= sin t. Drawing a vertical line at x= 1/2 I notice the vertical symmetry: [itex]cos\left(-\frac{\pi}{3}\right)= \frac{1}{2}[/itex] also. That gives me two solutions, between [itex]-\pi[/itex] and [itex]\pi[/tex]. If you want solutions between 0 and [itex]2\pi[/itex], [itex]-\frac{\pi}{3}+ 2\pi= \frac{5\pi}{3}[/itex] (NOT [itex]\frac{4\pi}{3}[/itex]).

    Since cosine is periodic with period [itex]2\pi[/itex], all solutions to 8cos x- 4= 0 are of the form [itex]\frac{\pi}{3}+ 2n\pi[/itex] or [itex]\frac{5\pi}{3}+ 2n\pi[/itex] for some integer n.
     
    Last edited: Aug 18, 2005
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