Use the sum to product identities
List of trigonometric identities - Wikipedia, the free encyclopedia
You should be able to finish from here
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
Regards,
Stonehambey
Use the sum to product identities
List of trigonometric identities - Wikipedia, the free encyclopedia
You should be able to finish from here
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
eek, sorry that was a typo on my part What I meant to type was
If we let n = 0 and sub back into the original equation we have
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
yeah, I accounted for that when I gave the general solution as
starting at and going up by each time gives
Starting at and going up by each time gives us
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)?