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Cos2x-6cosx-1=0 ,[0,π]

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asked Oct 7, 2018 in TRIGONOMETRY by anonymous

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1 Answer

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cos2x-6cosx-1=0

(2cos^2x-1)-6cosx-1=0

2cos^2x-6cosx-2=0

cos^2x-3cosx-1=0

cosx=(3+√13)/2 or( 3-√13)/2

x=107.46

answered Oct 9, 2018 by lilly Expert
reshown Oct 9, 2018 by bradely

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