SOLUTION: Show that cos 2pi/9 is a root of the equation 8x^3 - 6x + 1=0

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Question 1142053: Show that cos 2pi/9 is a root of the equation 8x^3 - 6x + 1=0
Found 2 solutions by math_helper, ikleyn:
Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!
I think you meant to say "Show it is NOT a root."

is a transcendental number - which means neither it, nor any rational multiple of it, is the root of ANY polynomial equation with integer coefficients.


It is the very definition of a transcendental number.
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EDIT: Oh, disregard, I just noticed cos in front, you should really write cos(2pi/9)
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cos(2pi/9) is approx 0.76604444
Plugging this into the equation gives 0 out, therefore cos(2pi/9) is a root.

Answer by ikleyn(52798)   (Show Source): You can put this solution on YOUR website!
.

            This problem has beautiful, nice, elegant and unexpected solution. 


Use the formula 


     =  - .


This formula is valid for any angle .


    For its proof see my post 

        - https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1142052.html 

    at this forum.


Let   =   and let  x = .


Notice that   =  =  =  = 120°.


Hence,   = .


From the other side,   =  - ,  according to the formula above.


In other words, 


     -  = .


Multiplying by 2 both sides and simplifying, you get


     -  +  = 0.


It means that   x =   is the solution of the given equation.

The proof is completed.



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