Results 1 to 5 of 5

Math Help - Solve the Following Trigonometric Equation

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    3

    Solve the Following Trigonometric Equation

    Hey, in pre-cal we're going over some of the stuff we learned this semester. Unfortunately for me, I have forgotten a bit of it:

    So, here's the problem:
    Solve the equation:
    sin2x + cos3x = 0 for [0, 2pi)

    I wish I could just graph it, but I have to do it algebraically and show all work. Any clues as to how to approach this would be appreciated

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,689
    Thanks
    448
    Quote Originally Posted by ayoYNr View Post
    Hey, in pre-cal we're going over some of the stuff we learned this semester. Unfortunately for me, I have forgotten a bit of it:

    So, here's the problem:
    Solve the equation:
    sin2x + cos3x = 0 for [0, 2pi)

    I wish I could just graph it, but I have to do it algebraically and show all work. Any clues as to how to approach this would be appreciated

    Thanks
    first task ... you'll need to expand \sin(2x) and \cos(3x)

    \sin(2x) = 2\sin{x}\cos{x} ... that's done.

    \cos(3x) = \cos(2x+x) = \cos(2x)\cos{x} - \sin(2x)\sin{x} = ... keep going
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    Posts
    3
    Ok, so now I have:

    2sinxcosx + cos(2x)cosx - sin(2x)sinx

    Where do I go from here?
    Last edited by ayoYNr; May 10th 2009 at 11:58 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2009
    Posts
    3
    (Sorry, ignore this post)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,689
    Thanks
    448
    Quote Originally Posted by ayoYNr View Post
    Ok, so now I have:

    2sinxcosx + cos(2x)cosx - sin(2x)sinx

    Where do I go from here?
    2\sin{x}\cos{x} + (1 - 2\sin^2{x})\cos{x} - 2\sin^2{x}\cos{x} = 0

    \cos{x}(2\sin{x} + 1 - 2\sin^2{x} - 2\sin^2{x}) = 0

    \cos{x}(1 + 2\sin{x} - 4\sin^2{x}) = 0

    \cos{x} = 0 ... should be straightforward to solve for x

    1 + 2\sin{x} - 4\sin^2{x} = 0 ... you'll need to use the quadratic formula to solve for \sin{x}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. solve for t in a trigonometric equation
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: May 5th 2009, 08:20 PM
  2. Solve the trigonometric equation?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 28th 2009, 08:15 AM
  3. Solve a trigonometric equation
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: March 8th 2009, 10:22 AM
  4. Solve the trigonometric equation
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: March 5th 2009, 08:08 AM
  5. trigonometric equation solve
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 30th 2008, 01:59 PM

Search Tags


/mathhelpforum @mathhelpforum