Pasegeabe85xy

2022-03-29

How to solve the cubic ${x}^{3}-3x+1=0$
This was a multiple choice question with options being
(A)$-\mathrm{cos}\frac{2\pi }{9},\mathrm{cos}\frac{8\pi }{9},\mathrm{cos}\frac{14\pi }{9}$
(B)$-2\mathrm{cos}\frac{2\pi }{9},2\mathrm{cos}\frac{8\pi }{9},2\mathrm{cos}\frac{14\pi }{9}$
(C)$-\mathrm{cos}\frac{2\pi }{11},\mathrm{cos}\frac{8\pi }{11},\mathrm{cos}\frac{14\pi }{11}$
(D)$-2\mathrm{cos}\frac{2\pi }{11},2\mathrm{cos}\frac{8\pi }{11},2\mathrm{cos}\frac{14\pi }{11}$
I tried to eliminate options using the sum and product of roots but I can't figure out if
$\mathrm{cos}\frac{2\pi }{9}+\mathrm{cos}\frac{8\pi }{9}+\mathrm{cos}\frac{14\pi }{9}=0$
or
$\mathrm{cos}\frac{2\pi }{11}+\mathrm{cos}\frac{8\pi }{11}+\mathrm{cos}\frac{14\pi }{11}=0$

membatas0v2v

Hint: ${\left(2\mathrm{cos}\theta \right)}^{3}-3\left(2\mathrm{cos}\theta \right)+1=2\mathrm{cos}3\theta +1$

Cason Harmon

Put x=2t
Hence
$8{t}^{3}-6t+1=0$
Now put $t=\mathrm{cos}\theta$ Hence
$2\mathrm{cos}\left(3\theta \right)=-1$
And the rest is simple trigonometric equation

Do you have a similar question?