Solutions of Polynomial Equations (1 Viewer)

nrlwinner · Mar 7, 2010

The Cmbridge book does not explain properly. Can someone guide me through the steps.

$cos3\Theta=4cos^3\Theta -3cos\Theta$

Solve for $8x^3-6x-1=0$

kcqn93 · Mar 7, 2010

@ is theta

let z = cis@

using demoivre's theorem find z^3

then using binomials or pascal triangle find z^3

equate for real and imaginary for both equations

you should get the answer

im not sure about the rest lol sorry

nrlwinner · Mar 7, 2010

kcqn93 said:
@ is theta

let z = cis@

using demoivre's theorem find z^3

then using binomials or pascal triangle find z^3

equate for real and imaginary for both equations

you should get the answer

im not sure about the rest lol sorry

Yeah I already know about that part. But it's the actually solution I don't know how to get.

cutemouse · Mar 7, 2010

nrlwinner said:
The Cmbridge book does not explain properly. Can someone guide me through the steps.

$cos3\Theta=4cos^3\Theta -3cos\Theta$

Solve for $8x^3-6x-1=0$

$cos3\theta = 4cos^3\theta-3cos\theta \\ Let \ cos\3\theta= 1/2 \ \3\theta = \frac{\pi}3 , \ \frac{5\pi}{3} \ or \ \frac{7\pi}{3} \\ \theta = \frac{\pi}9 , \ \frac{5\pi}{9} \ or \ \frac{7\pi}{9} \\ Thus \ 4cos^3\theta - 3cos\theta = \frac{1}{2} \text{ will have the same roots} \\ Let \ x=cos\theta \ \therefore 4x^3-3x=\frac{1}{2} \\ \therefore \ 8x^3-6x-1=0 \text{ has the roots } x=cos\frac{\pi}9, \ cos\frac{5\pi}{9} \ or \ cos\frac{7\pi}{9}$

nrlwinner · Mar 7, 2010

jm01 said:
$cos3\theta = 4cos^3\theta-3cos\theta \\ Let \ cos\3\theta= 1/2 \ \3\theta = \frac{\pi}3 , \ \frac{5\pi}{3} \ or \ \frac{7\pi}{3} \\ \theta = \frac{\pi}9 , \ \frac{5\pi}{9} \ or \ \frac{7\pi}{9} \\ Thus \ 4cos^3\theta - 3cos\theta = \frac{1}{2} \text{ will have the same roots} \\ Let \ x=cos\theta \ \therefore 4x^3-3x=\frac{1}{2} \\ \therefore \ 8x^3-6x-1=0 \text{ has the roots } x=cos\frac{\pi}9, \ cos\frac{5\pi}{9} \ or \ cos\frac{7\pi}{9}$

Why does cos3@=1/2 ?

Timothy.Siu · Mar 7, 2010

nrlwinner said:
Why does cos3@=1/2 ?

coz thats what you do to solve the question.
u want an equation in the form of 8x^3-6x-1=0
so u let cos 3@=1/2.
its merely to solve the equation.
if it was say, another number instead of -1, u would sub in something else.

untouchablecuz · Mar 7, 2010

nrlwinner said:
The Cmbridge book does not explain properly. Can someone guide me through the steps.

$cos3\Theta=4cos^3\Theta -3cos\Theta$

Solve for $8x^3-6x-1=0$

$\\8x^3-6x-1=0\\4x^3-3x-\frac{1}{2}=0\\4x^3-3x=\frac{1}{2}\\$Let $ x $ be of the form $\cos \theta,\\$Hence, $ 4\cos^3\theta-3\cos \theta=\frac{1}{2}\\$But we are given that $\cos 3\theta=4\cos^3\theta-3\cos \theta\\\therefore \cos 3\theta=\frac{1}{2}\\3\theta=2 \pi k+\cos^{-1}\left(\frac{1}{2}\right)\\\theta=\frac{\pi}{9}(6k+1) \ \ \forall k\in\mathbb{Z} \\\therefore x=\cos \theta =\cos \left(\frac{\pi}{9}(6k+1)\right)\\$From the fundamental theorem of algebra, there exists three unique solutions to the equation in $x\\$Namely, $k=0,1,-1\\\therefore x=\cos\frac{\pi}{9},\cos\frac{7\pi}{9} $ and $\cos\frac{-5\pi}{9} (=\cos\frac{5\pi}{9} $ since $\cos(-x)=\cos(x))$

cutemouse · Mar 7, 2010

nrlwinner said:
Why does cos3@=1/2 ?

You let it equal 1/2 for the required expression.

Solutions of Polynomial Equations (1 Viewer)

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