Results 1 to 2 of 2

Math Help - Solutions between (0,2pi]

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    5

    Solutions between (0,2pi]

    I am just trying to get some help on how to get this equation right. I have it equal to 0. Anything will help thanks.

    sin5x-sinx=2cos3x
    sin4x-2cos3x=0 (I think this is correct, just need help moving on to the next step.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by PintsofGuinness View Post
    sin5x-sinx=2cos3x
    sin4x-2cos3x=0

    What?.

    \sin A-\sin B=2\cos\dfrac{A+B}{2}\sin \dfrac{A-B}{2}

    So,

    \sin 5x-\sin x=2\cos 3x \sin 2x
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 4th 2011, 09:21 PM
  2. solutions
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 11th 2010, 06:07 AM
  3. Replies: 6
    Last Post: July 26th 2010, 12:45 PM
  4. help with ODE. *i got no solutions
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: October 10th 2009, 10:12 AM
  5. Solutions to ODE y'(t) = t^2 +y(t)^2
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: January 18th 2009, 01:44 AM

Search Tags


/mathhelpforum @mathhelpforum