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Complex number question involving de Moivre identity

  1. Jan 10, 2012 #1
    1. The problem statement, all variables and given/known data

    cos(4x)(6+2a)+12a+8b=-20 find values for a, b. Then check the values and state which values of x would not have been sufficient checks.

    2. Relevant equations

    Complex number equations

    3. The attempt at a solution

    I've simplified it down to this from a harder problem but I can't get any further, putting it into wolfram gives me a=-3 and b=2 but I have no idea how that was worked out.

    Thanks in advance for any help!
  2. jcsd
  3. Jan 10, 2012 #2
    Every value for x you plug in gives you an equation that a and b must satisfy.

    So, why don't you pick some easy values for x and see what you get??
  4. Jan 10, 2012 #3
    Since cos(4x) is varying independently, then in order to have the equation always true, what value must (6+2a) have?
  5. Jan 11, 2012 #4
    Yeah my fault, I didn't say that you have to do it using the de Moivre identity and then plug in a value of x to test it afterwards.

    The original equation looks like this: [itex]cos(x)^4 + sin(x)^4 + a(cos(x)^2 + sin(x)^2) + b = 0[/itex]

    I think it's a fairly common complex number question and involves the bionomial expansion, but I've never really done much with complex numbers before so it doesn't seem obvious what I need to do to me. Thanks!
  6. Jan 12, 2012 #5
    Alright I've figured out how to get a and b, anyone have any ideas on which values of x would not be 'sufficient checks' on the formula?
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