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Trig problem cos4x-7cos2x=8

  1. Aug 17, 2008 #1
    1. The problem statement, all variables and given/known data

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

    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[tex]^{}2[/tex]3x-sin[tex]^{}2[/tex]3x. So from that I can assume that cos4x= cos[tex]^{}2[/tex]2x-sin[tex]^{}2[/tex]2x. But how would cos2x=2cos[tex]^{}2[/tex]x-1 change when its cos4x? I got my problem down to:

    cos4x-14cos[tex]^{}2[/tex]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. jcsd
  3. Aug 17, 2008 #2
    Ok well let's say we have [tex] cos(2y) = 2cos^2(y) - 1 [/tex]. Now to get [tex]cos(4x)[/tex], let [tex]y=2x[/tex]. 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.
     
  4. Aug 17, 2008 #3
    I get what you are saying, but I still cannot solve the problem. :confused:
     
  5. Aug 17, 2008 #4
    You have the right idea. Write everything in terms of [tex]cos(2x)[/tex]and constants. Then see if you move everything one side and factor.
     
  6. Aug 17, 2008 #5
    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 [tex]\pi[/tex]/2 and 3[tex]\pi[/tex]/2 implying that the answer is cosx=0 unless I am just having issues with my calculator which happens from time to time.

    thanks
     
  7. Aug 18, 2008 #6

    dynamicsolo

    User Avatar
    Homework Helper

    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!)
     
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