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Thread: General Solution

  1. #1
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    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.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    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.
    $\displaystyle \cos(5x)=\sin(4x)$

    Interesting...

    well, you had the right idea at first:

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

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

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

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

    the notation $\displaystyle n\in\mathbb{Z}$ 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
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    $\displaystyle \cos(5x)=\sin(4x)$

    Interesting...

    well, you had the right idea at first:

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

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

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

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

    the notation $\displaystyle n\in\mathbb{Z}$ 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 $\displaystyle \sin \alpha = \sin \beta$ are $\displaystyle \alpha = \beta + 2 n \pi$ (the result Chris has used) AND $\displaystyle \alpha = \pi - \beta + 2 n \pi$ ....

    You should derive the general form of the remaining solutions: $\displaystyle x = \frac{(4n-1)}{18} \pi$ where n has been previously defined.
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  4. #4
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    Hello, guest12345!

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


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

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

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

    Get it?

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


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

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


    (I'll let you generalize this answer.)
    .
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