# Complex number question involving de Moivre identity

1. Jan 10, 2012

### Stickybees

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. Jan 10, 2012

### micromass

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??

3. Jan 10, 2012

### Joffan

Since cos(4x) is varying independently, then in order to have the equation always true, what value must (6+2a) have?

4. Jan 11, 2012

### Stickybees

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: $cos(x)^4 + sin(x)^4 + a(cos(x)^2 + sin(x)^2) + b = 0$

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!

5. Jan 12, 2012

### Stickybees

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?