Complex number question involving de Moivre identity

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Homework Statement



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.

Homework Equations



Complex number equations

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!
 
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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??
 
Since cos(4x) is varying independently, then in order to have the equation always true, what value must (6+2a) have?
 
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!
 
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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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