Let `n` be a positive integer such that `sinpi/(2n)+cospi/(2n)=(sqrt(n))/2dot` Then `6lt=nlt=8` (b) `4

Explanation

We have,
`sin pi/(2n) + cos(pi/(2n))=sqrt(n)/2`
Squaring both sides, we get
`sin^(2) pi/(2n) + cos^(2) pi/(2n) + 2 sin pi/(2n) .cos pi/(2n)=n/4`
`rArr 1+sin2(pi/(2n))=n/4`
`rArr sin(pi/n) = n/4-1 = (n-4)/4`
For `n=2, sinpi/2 =1` and `(2-4)/4 = 1/2`
`rArr n=2` does not satisfy the given equation
Considering `n gt 1` and `n ne 2, 0 lt sin pi/n lt 1`
`rArr 0 lt (n-4)/4 lt 1`
`rArr 4 lt n lt 8`.