# Homework Help: Trig problem cos4x-7cos2x=8

1. Aug 17, 2008

### Geekchick

1. The problem statement, all variables and given/known data

cos4x-7cos2x=8 solve in interval [0,2$$\pi$$)

2. Relevant equations

reciprocal identities, pythagorean identities, sum/difference formulas, double angle formulas, power reducing formula, half-angle formula, product to sum formula, sum to product formula.

3. The attempt at a solution

So to solve this problem I am pretty sure I will have to use the double angle formula. However, I am a little confused as to how the formula changes when you have more than 2x. the only additional example my textbook gives it Cos6x=cos$$^{}2$$3x-sin$$^{}2$$3x. So from that I can assume that cos4x= cos$$^{}2$$2x-sin$$^{}2$$2x. But how would cos2x=2cos$$^{}2$$x-1 change when its cos4x? I got my problem down to:

cos4x-14cos$$^{}2$$x-15

I'm hoping that if I can expand on cos4x then I can maybe factor...or am I going about this all wrong.
Please help, I hate to be a bother but I'm taking this Pre-cal class independently over the summer and have already got an incomplete grade. My teacher said I have until the 20th to finish all the test and labs, so I'm trying to tie up all my loose ends so I can be done and move onto calculus.
Thanks

2. Aug 17, 2008

### snipez90

Ok well let's say we have $$cos(2y) = 2cos^2(y) - 1$$. Now to get $$cos(4x)$$, let $$y=2x$$. It should make sense now. I used to have the same problem, which kind of goes back to functions. But if you think about it a little it shouldn't be hard to work out.

3. Aug 17, 2008

### Geekchick

I get what you are saying, but I still cannot solve the problem.

4. Aug 17, 2008

### snipez90

You have the right idea. Write everything in terms of $$cos(2x)$$and constants. Then see if you move everything one side and factor.

5. Aug 17, 2008

### Geekchick

Ok, so this is what I have done.

cos4x-7cos2x=8

seeing as cos4x=2cos22x-1

then

2cos22x-7co2x-9

When I factor that I get

(2cos2x-9)(cos2x+1)

Right so far?

then when I set each to zero I get

2cos2x=9

Here is where I'm a little confused. Does it go like this?

cos2x=9/2

cosx=9/4

or is it

cosx=9/2

all of these twos are throwing me lol

With the other one I get

cos2x=1
cosx=1/2

But another issue is when I plug the problem into my calculator and graph I end up with the answer being $$\pi$$/2 and 3$$\pi$$/2 implying that the answer is cosx=0 unless I am just having issues with my calculator which happens from time to time.

thanks

6. Aug 18, 2008

### dynamicsolo

OK up to here...

This is correct. And this should instantly tell you that this case has no solution, as there is no angle (doubled or otherwise) which has a cosine larger than 1...

Neither of these is correct. It is not the case that cos(nx) = n cos(x) ; you cannot simply pull constant factors out of the argument of a trig function -- those constants are multiplying the angle and will not similarly multiply the trig function.

There is also a problem here. Your result for this case is (cos2x+1) = 0 , which gives

cos 2x = -1.

Again, you cannot just pull that '2' out of the argument of cosine: it is not something called cosine times 2x, but a function called cosine applied to the angle (2x).

The way you solve this is to ask what angles have a cosine of -1 (equivalently, we would solve 2x = arccosine(-1) ). We need all the angles in the interval [0, 2·pi), so we would look at

2x = (pi) , 2x = 3(pi), 2x = 5(pi), ... ,

which gives us

x = (pi)/2 , x = 3(pi)/2 [and that's it: everything beyond that is outside the specified domain]. Since the other factor gave no other solutions, these are the only ones.

As a check (which you should always do when working with trig equations), we find

cos4x - 7cos2x = 8

x = (pi)/2 : cos(2·pi) - 7·cos(pi) = 1 - 7·(-1) = 8 (check!)

x = 3(pi)/2 : cos(6·pi) - 7·cos(3·pi) = 1 - 7·(-1) = 8 (check!)