Results 1 to 5 of 5

Math Help - Roots of sec(x) function with transformation

  1. #1
    Junior Member
    Joined
    Jun 2011
    Posts
    38

    Roots of sec(x) function with transformation

    I was able to solve a similar question finding the roots of the tan(x) function with a transformation, but this one seems to be giving me a bit more trouble...

    The equation is y = 2sec(-2x + 180) + 3

    I know to find the roots I have to set y to 0 and solve for x.

    This is what I've done so far:

    y = 2sec(-2x + 180) + 3
    0 = 2sec (-2x + 180) + 3
    -3 = 2sec (-2x + 180)
    -3/2 = sec (-2x + 180)
    sec^-1 (-3/2) = -2x + 180

    This is where I'm running into trouble. I know that sec = 1/cos, so I rewrote this line as:

    1/(cos^-1 (-3/2) ) = -2x + 180

    But I'm not able to solve for cos^-1 (-3/2) so now I'm at a bit of a loss.

    Any help would be greatly appreciated. Thanks so much!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by starshine84 View Post
    I was able to solve a similar question finding the roots of the tan(x) function with a transformation, but this one seems to be giving me a bit more trouble...

    The equation is y = 2sec(-2x + 180) + 3

    I know to find the roots I have to set y to 0 and solve for x.

    This is what I've done so far:

    y = 2sec(-2x + 180) + 3
    0 = 2sec (-2x + 180) + 3
    -3 = 2sec (-2x + 180)
    -3/2 = sec (-2x + 180)
    Correct up to here.

    sec^-1 (-3/2) = -2x + 180

    This is where I'm running into trouble. I know that sec = 1/cos, so I rewrote this line as:

    1/(cos^-1 (-3/2) ) = -2x + 180

    But I'm not able to solve for cos^-1 (-3/2) so now I'm at a bit of a loss.
    You need to take the reciprocal of both sides (you only did the right hand side)

    -\dfrac{3}{2} = \sec (-2x + 180) \rightarrow -\dfrac{2}{3} = \cos (-2x+180)
    Last edited by e^(i*pi); June 13th 2011 at 07:04 AM. Reason: tex
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2011
    Posts
    38
    Okay, so from there, is this right?

    -2/3 = cos(-2x + 180)
    cos^-1 (-2/3) = cos(-2x + 180)
    -48.1896851 = -2x
    24.09484255 = x
    Follow Math Help Forum on Facebook and Google+

  4. #4
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    You can test potential solutions by seeing if they work in the original equation.

    This one is a root but it's only the principle root - since \sec(x) is periodic you're going to have infinite roots unless the question specifies a range (0 \leq x \leq 360) is common
    Last edited by e^(i*pi); June 13th 2011 at 09:29 AM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jun 2011
    Posts
    38
    Okay, I think I'm on the right track now. Thanks for you help!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. transformation of an exponential function
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 2nd 2010, 09:44 AM
  2. Replies: 1
    Last Post: December 3rd 2009, 08:45 AM
  3. function transformation
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: September 26th 2008, 08:53 PM
  4. A Function Transformation
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 9th 2008, 03:17 PM
  5. function transformation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 28th 2005, 02:04 AM

Search Tags


/mathhelpforum @mathhelpforum