Problem of the week #63 - June 9th, 2013

[Jameson]

Thank you to MarkFL for proposing this problem for us to use![HR][/HR]
Using the triple-angle identity for cosine:

(1) $\displaystyle \cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)$

solve the cubic equation:

(2) $\displaystyle 8x^3-6x-1=0$
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Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Jameson]

Congratulations to the following members for their correct solutions:

1) anemone
2) kaliprasad
3) Sudharaka

Solution (from Sudharaka):

Spoiler

Substitute \(x=\cos\theta\) in the cubic equation. Then we get,

\[8\cos^3\theta-6\cos\theta-1=0\]

\[\Rightarrow 2(4\cos^3\theta-3\cos\theta)=1\]

Using the Triple angle identity for cosine we get,

\[\Rightarrow \cos(3\theta)=\frac{1}{2}\]

Hence, \(\displaystyle 3\theta=\frac{\pi}{3}, \frac{7\pi}{3}, \frac{13\pi}{3}\) are three distinct values which satisfy the given cubic equation.

\[\therefore \theta=\frac{\pi}{9}, \frac{7\pi}{9}, \frac{13\pi}{9}\]

So, the roots of the cubic equation are,

\[x= \cos{ \left( \frac{\pi}{9} \right)},\, \cos{ \left( \frac{7\pi}{9} \right)},\, \cos{\left( \frac{13\pi}{9} \right)}\]