One interesting problem,some help? "If a+bx+cx^2=0 where x=cos 20..."

[Andrei19997]

Hello!
The problem sound like this:
If a,b,c are rational numbers and a+bx+cx^2=0 where x=cos 20 degrees;Prove that a=b=c=0.

 

[Deleted member 4993]

Hello!
The problem sound like this:
If a,b,c are rational numbers and a+bx+cx^2=0 where x=cos 20 degrees;Prove that a=b=c=0.

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

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[Andrei19997]

I don't have any idea how I should start;I just want to know and learn how to solve this kind of problem.Maybe a little help in the begining...

 

[Steven G]

I don't have any idea how I should start;I just want to know and learn how to solve this kind of problem.Maybe a little help in the begining...

Have you tried to replace x with cos 20 degrees? What does cos 20 degrees equal? What do you get when you square it. This is what I would try first. Let's see where that gets us.

 

[Andrei19997]

I don't get it.How do i square cos 20 ?What form does cos 20 have? The one from wolfram alfa?

 

[Ishuda]

Hello!
The problem sound like this:
If a,b,c are rational numbers and a+bx+cx^2=0 where x=cos 20 degrees;Prove that a=b=c=0.

Would it be easier if a, b, and c were integers with a greatest common divisor of 1?