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The number of distinct solutions of the equation `5/4cos^(2)2x + cos^4 x + sin^4 x+cos^6x+sin^6 x =2` in the interval `[0,2pi] ` is

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Correct Answer - -8
Here, `(5)/(4) cos^(2)2x + (cos^(4)x +sin^(4)x) + (cos^(6)x + sin^(6)x) =2`
`rArr (5)/(4) cot2x +[(cos^(2)x + sin^(2) x)^(2) -2sin^(2) x cos^(2)x]+(cos ^(2)x + sin^(2)x) [(cos^(2)x + sin^(2)x)^(2)-3sin^(2) cos^(2) x]=2`
`rArr (5)/(4)cos^(2)x + (1-2sin^(2)x cos^(2)x) + (1-3cos^(2) xsin^(2)x)=2`
`rArr (5)/(4)cos^(2)2x-5sin^(2)x cos^(2)x = 0`
`rArr (5)/(4)cos^(2)2x-(5)/(4) sin^(2)2x=0`
`rArr (5)/(2)cos^(2)2x = (5)/(4)rArr cos^(2) 2x= (1)/(2)`
`rArr 2cos^(2)2x =1`
`rArr 1+ cos 4x =1`
`rArr cos 4x =0 as 0 le x le 2pi`
`therefore 4x={(pi)/(2),(3pi)/(8),(5pi)/(2),(7pi)/(2),(9pi)/(2),(11pi)/(2),(13pi)/(2),(15pi)/(2)}`
as ` 0 le 4x le 8 pi`
`rArr x = {(pi)/(8),(3pi)/(8),(5pi)/(8),(9pi)/(8),(11pi)/(8),(13pi)/(8),(15pi)/(8)}`
Hence ,the total number of solutions is 8.

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