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roots of complex number

20160717_121548[1].jpgHow to solve for the roots part for question 19?
(edited 7 years ago)
set z=cosx

hence the polynomial in Z has solutions(6 unique solutions) when cos6x=0, which is easy to find solutions for.
wat modules this
Original post by magicelephant
wat modules this


In OCR MEI this would be FP2.
Reply 4
cos6x=0x=±π12+nπ3,nZ\displaystyle \cos 6x=0 \Rightarrow x=\pm \frac{\pi}{12} + \frac{n\pi}{3} , n \in \mathbb{Z} .
(edited 7 years ago)

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