Results 1 to 4 of 4

Math Help - General Solution

  1. #1
    Newbie
    Joined
    Aug 2008
    Posts
    7

    General Solution

    Can someone help me with this equation please?
    Find the general solution to : cos(5x) = sin(4x)
    I had it for a test today and just want to know the answer coz I had no idea what to do
    I started off changing it to sin(pi/2 - 5x) - sin(4x) but not sure if it's right and if it is not sure where to go from here.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by guest12345 View Post
    Can someone help me with this equation please?
    Find the general solution to : cos(5x) = sin(4x)
    I had it for a test today and just want to know the answer coz I had no idea what to do
    I started off changing it to sin(pi/2 - 5x) - sin(4x) but not sure if it's right and if it is not sure where to go from here.
    \cos(5x)=\sin(4x)

    Interesting...

    well, you had the right idea at first:

    let \cos(5x)=\sin\left(\tfrac{\pi}{2}-5x\right)

    Thus, \sin\left(\tfrac{\pi}{2}-5x\right)=\sin(4x)

    For this equation to be true, \tfrac{\pi}{2}-5x=4x

    Note that there is no given restriction on x. So we can generalize this and say that this equation is true when \tfrac{(4n+1)}{2}\pi=9x\implies\color{red}\boxed{x  =\tfrac{(4n+1)}{18}\pi; \ n\in\mathbb{Z}}

    the notation n\in\mathbb{Z} means that n is an element of the integer set (this lets us know that n can be ...-3, -2, -1, 0, 1, 2, 3,...).

    I hope this makes sense!

    --Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Chris L T521 View Post
    \cos(5x)=\sin(4x)

    Interesting...

    well, you had the right idea at first:

    let \cos(5x)=\sin\left(\tfrac{\pi}{2}-5x\right)

    Thus, \sin\left(\tfrac{\pi}{2}-5x\right)=\sin(4x)

    For this equation to be true, \tfrac{\pi}{2}-5x=4x

    Note that there is no given restriction on x. So we can generalize this and say that this equation is true when \tfrac{(4n+1)}{2}\pi=9x\implies\color{red}\boxed{x  =\tfrac{(4n+1)}{18}\pi; \ n\in\mathbb{Z}}

    the notation n\in\mathbb{Z} means that n is an element of the integer set (this lets us know that n can be ...-3, -2, -1, 0, 1, 2, 3,...).

    I hope this makes sense!

    --Chris
    Chris has provided half of the solutions .....

    Note that the solutions to \sin \alpha = \sin \beta are \alpha = \beta + 2 n \pi (the result Chris has used) AND \alpha = \pi - \beta + 2 n \pi ....

    You should derive the general form of the remaining solutions: x = \frac{(4n-1)}{18} \pi where n has been previously defined.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,738
    Thanks
    644
    Hello, guest12345!

    There is a back-door approach to this problem . . .


    Find the general solution to: . \cos(5x) \:= \:\sin(4x)

    Look at this diagram.
    Code:
                            *
                         *  *
                  c   *  5x *
                   *        * a
                *           *
             * 4x           *
          *  *  *  *  *  *  *
                    b

    We have: . \begin{array}{ccc}\sin(4x) &=&\frac{a}{c} \\ \cos(5x) &=&\frac{a}{c} \end{array}

    Get it?

    If \sin(A) = \cos(B), then A and B are in the same right triangle.
    . . The angles are complementary: . A + B \:=\:\frac{\pi}{2}


    Hence: . 5x + 4x \:=\:\frac{\pi}{2} \quad\Rightarrow\quad 9x \:=\:\frac{\pi}{2}

    . . Therefore: . x \:=\:\frac{\pi}{18}


    (I'll let you generalize this answer.)
    .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. General Solution of a differential solution
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: September 11th 2010, 02:49 AM
  2. General Solution
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: May 21st 2010, 04:05 AM
  3. Finding the general solution from a given particular solution.
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: October 7th 2009, 01:44 AM
  4. find the general solution when 1 solution is given
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: March 4th 2009, 09:09 PM
  5. DE - General Solution
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: February 1st 2009, 03:55 PM

Search Tags


/mathhelpforum @mathhelpforum