general solution to trig problem

• May 17th 2009, 10:47 AM
Stonehambey
general solution to trig problem
Hi,

I'm having a lot of trouble answering this question it must be said, I've tried a number of approaches, mostly using the double angle formulae on sin4x and cos4x and then messing around with it to get it to look like something I can solve, but to no avail. Any help would be much appreciated! :)

Find the general solution to

$\sin 2x + \sin 4x = \cos 2x + \cos 4x$

Regards,

Stonehambey
• May 17th 2009, 11:56 AM
TheEmptySet
Quote:

Originally Posted by Stonehambey
Hi,

I'm having a lot of trouble answering this question it must be said, I've tried a number of approaches, mostly using the double angle formulae on sin4x and cos4x and then messing around with it to get it to look like something I can solve, but to no avail. Any help would be much appreciated! :)

Find the general solution to

$\sin 2x + \sin 4x = \cos 2x + \cos 4x$

Regards,

Stonehambey

Use the sum to product identities

List of trigonometric identities - Wikipedia, the free encyclopedia

$2\sin\left(\frac{2x+4x}{2} \right)\cos\left(\frac{2x-4x}{2} \right)=2\cos\left(\frac{2x+4x}{2} \right)\cos\left(\frac{2x-4x}{2} \right)$

$2\sin(3x)\cos(-2x)=2\cos(3x)\cos(-2x)$

$2\sin(3x)\cos(2x)=2\cos(3x)\cos(2x)$

$2\sin(3x)\cos(2x)-2\cos(3x)\cos(2x)=0$

$2\sin(3x)\cos(2x)-2\cos(3x)\cos(2x)=0$

$2\cos(2x)[\sin(3x)-\cos(3x)]=0$

You should be able to finish from here
• May 17th 2009, 12:43 PM
Stonehambey
sum to product! Now it's obvious :P

Thanks! :)
• May 17th 2009, 12:47 PM
Calculus26
I'm assuming you are looking for all solutions on interval [0,2pi)

I include thr graph just to show we can expect 8 solutions

I've included the solutions and method in the attachment .

But the basic method i used was to rearrange as

sin(2x)-cos(2x) = cos(4x)-sin(4x)

Then I squared both sides and got 2 fairly simple equations

however there are 16 possibilities of which 8 are solutions--this is the tedious part.

If someone has an easier way I'd like to know it.

should have waited
• May 17th 2009, 01:07 PM
Stonehambey
hmm, I tried the way theEmptySet suggested and got a little stuck again (Crying)

I applied the formulas and got to the same bit that you did. When I solved for

$2\cos(2x)=0$

I get $x=\frac{\pi}{4} + n\pi$, which doesn't work when I sub back into the original equation, so where did I go wrong? :)
• May 17th 2009, 01:12 PM
TheEmptySet
Quote:

Originally Posted by Stonehambey
hmm, I tried the way theEmptySet suggested and got a little stuck again (Crying)

I applied the formulas and got to the same bit that you did. When I solved for

$2\cos(2x)=0$

I get $x=\frac{\pi}{2} + n\pi$, which doesn't work when I sub back into the original equation, so where did I go wrong? :)

You are really close you should have gotten

$2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}$
• May 17th 2009, 01:26 PM
Stonehambey
Quote:

Originally Posted by TheEmptySet
You are really close you should have gotten

$2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}$

eek, sorry that was a typo on my part :) What I meant to type was

$\cos(2x) = 0$

$2x = \pm \frac{\pi}{2} + 2\pi n$

$\implies x = \pm \frac{\pi}{4} + n \pi$

If we let n = 0 and sub back into the original equation we have

$\sin\left(\frac{\pi}{2}\right) + \sin(\pi) = \cos\left(\frac{\pi}{2}\right) + \cos(\pi)$

$1 = -1$

which is obviously wrong but I can't see where the error is :)

EDIT: I got Calculus26 method to work, very nice idea to square both sides I didn't think of that. Now I'm wondering why I can't get it to work using the sum to product way =P
• May 17th 2009, 01:29 PM
TheEmptySet
Quote:

Originally Posted by Stonehambey
eek, sorry that was a typo on my part :) What I meant to type was

$\cos(2x) = 0$

$2x = \pm \frac{\pi}{2} + 2\pi n$

$\implies x = \pm \frac{\pi}{4} + n \pi$

If we let n = 0 and sub back into the original equation we have

$\sin\left(\frac{\pi}{2}\right) + \sin(\pi) = \cos\left(\frac{\pi}{2}\right) + \cos(\pi)$

$1 = -1$

which is obviously wrong but I can't see where the error is :)

EDIT: I got Calculus26 method to work, very nice idea to square both sides I didn't think of that. Now I'm wondering why I can't get it to work using the sum to product way =P

Cosine has a zero every Pi units 2pi is too much see my above post.
• May 17th 2009, 01:34 PM
TheEmptySet
Quote:

Originally Posted by TheEmptySet
Cosine has a zero every Pi units 2pi is too much see my above post.

You are right it only works for odd n.
• May 17th 2009, 02:35 PM
Stonehambey
Quote:

Originally Posted by TheEmptySet
Cosine has a zero every Pi units 2pi is too much see my above post.

yeah, I accounted for that when I gave the general solution as

$2x = \pm\frac{\pi}{2} + 2n\pi$

starting at $\frac{\pi}{2}$ and going up by $2\pi$ each time gives $\frac{\pi}{2}, \frac{5\pi}{2}, \frac{9\pi}{2}, ...$

Starting at $-\frac{\pi}{2}$ and going up by $2\pi$ each time gives us $\frac{3\pi}{2}, \frac{7\pi}{2}, \frac{11\pi}{2}, ...$

So all the zeros are covered by these two sets of solutions :)

(obviously n could be negative and we get zeros going the other way as well)

But how do I know which ones will hold true for the original equation (when I divide them by 2 to get x on its own)?
• May 18th 2009, 02:28 PM
Stonehambey
Quote:

Originally Posted by TheEmptySet
You are really close you should have gotten

$2x = \frac{\pi}{2}+n\pi \iff x =\frac{\pi}{4}+n\frac{\pi}{2}$

So can any one explain why this only works for odd n? This problems is driving me up the wall!! (Headbang)

Regards,

Stonehambey