# Math Help - General Solution

1. ## 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.

2. Originally Posted by guest12345
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

3. Originally Posted by Chris L T521
$\cos(5x)=\sin(4x)$

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.

4. 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.)
.