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Finding the Zeroes of a Secant Graph

  1. Oct 13, 2008 #1
    This is a continuation of my previous post, so please bear with me

    1. The problem statement, all variables and given/known data
    Find the zeroes of the following function:

    y= 2 sec (-2x+180deg) + 3


    2. Relevant equations

    Break down the equation into:

    y= 2 sec -2(x-90deg) + 3;

    Finding zeroes, means finding the x values when y=0, therefore:

    0=2 sec [-2(x-90deg)]+3



    3. The attempt at a solution

    0=2 sec [-2(x-90deg)]+3

    -3 = 2 sec [-2(x-90deg)]

    -1.5 = sec [-2(x-90deg)]

    -1.5/sec = [-2(x-90deg)]

    ^ this is where I get stuck. Because the inverse of sec or cos (sec^-1; cos^-1) is Error or Undefined. So the whole thing explodes. Now, I can tell when the graph crosses the x-axis (and therefore has a y=0 value) on a Graphing tool, but how can I find this out algebraically?

    Thanks for the assistance.
     
    Last edited: Oct 13, 2008
  2. jcsd
  3. Oct 13, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Now that's just silly. "sec" is NOT a number, sec[-29x- 90)] is NOT just "sec" times [-29x- 90]. You cannot just divide by "sec". You get rid of the secant function by using its inverse, arcsec(x). -2(x- 90)= arcsec(-1.5). Since secant of 1/cosine and cosine is always between - 1 and 1, secant= 1/cosine is between -1 and -infinity or 1 and infinity. arcsec(x)= sec-1(x) is definitely defined for x= -1.5.

     
  4. Oct 13, 2008 #3
    editted
     
    Last edited: Oct 14, 2008
  5. Oct 13, 2008 #4
    This is what I am getting:
    -2(x- 90)= arcsec(-1.5)

    -2(x- 90)=0.999657325
    x-90=-0.499828662
    x=89.5
     
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