If a root of the equation `n^(2)sin^(2)x-2sinx-(2n+1)=0` lies in`[0,pi//2]` the minimum positive integer value of `n` is

Explanation

`:'n^(2)sin^(2)x-2sinx-(2n+1)=0`
`impliessinx=(2+-sqrt(4+4n^(2)(2n+1)))/(2n^(2))` [by Shridharacharya method]
`=(1+-sqrt((2n^(3)+n^(2)+1)))/(n^(2))`
`:'0 le sin x le 1 [:' x epsilon[0,pi//2]`
`implies0le(1+sqrt((2n^(3)+n^(2)+1)))/(n^(2))le1`
`implies0le1+sqsrt((2n^(3)+n^(2)+1))len^(2)`
`impliessqrt((2n^(3)+n^(2)+1))le(n^(2)-1) [ :' n gt1]`
ON squaring both sides we get
`2n^93)+n^(2)+1len^(4)-2n^(2)+1`
`impliesn^(4)-2n^(3)-3n^(2)ge0`
`impliesn^(2)-2n-3ge0implies(n-3)(n+1)ge0`
`impliesnge3`
`:.n=3,4,5`, ltbr. Hence the minimum positive integer value of `n` is 3.