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Trig Typo?

  1. Oct 31, 2008 #1
    b]1. The problem statement, all variables and given/known data[/b]
    Hey, I have a problem that I worked out in a trigonometry book. I got different answers from what the book gave, but I think that I am right and the book is wrong. Could someone please tell me who is right?

    2. Relevant equations

    Okay, they gave me the following problem: solve for x in secx^2-4=0 when 0 is less than or equal to x, which is less than or equal to 360.

    Here is how the book solves it:

    secx-2=0 secx+2=0
    secx=2 secx=-2
    x=sec^-1(2) x=sec^-1(-2)
    = 30, 330 = 150, 210

    Okay, here's where I have a big problem: secant is the reciprocal function of cosine, right?
    So if we evaluate x=sec^-1(2), we get cos^-1(1/2), which does not equal 30 and 330 when plugged into the calculator. Same situation with x=sec^-1(-2).
    I am pretty sure that this is a horrible typo, but I would be delighted to be corrected otherwise.

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Oct 31, 2008 #2
    I agree with you. If you want places where cos(x) = +/-.5 then that is 60,120,240,300. Also I graphed [sec(x)]^2-4 and it was 0 at those places. The values they gave were correct for sine. So unless the original problem is [csc(x)]^2-4=0, their answer is wrong.
  4. Oct 31, 2008 #3


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    Actually sec x = 1/cos x. I think you are confusing ^-1 and inverse functions.
  5. Oct 31, 2008 #4
    I think that there is a typo, the answers should actually be x = 60, 120, 240, 300.

    Solution from where the book leaves off:
    sec(x) = 2, sec(x) = -2
    1/2 = cos(x), -1/2 = cos(x)
    Use graph or calculator to find the angles are those that I stated above.
  6. Nov 1, 2008 #5


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    Staff Emeritus
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    No, he said "reciprocal", not "inverse". There is NO cos-1(2).
  7. Nov 1, 2008 #6


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    Gold Member

    Oops. Sorry.
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