Find the value of `x` in the interval `[-pi/2,(3pi)/2\ ]` for which `sqrt(2)sin2x+1lt=2sinx+sqrt(2\ )cosx`

Explanation

We have, `sqrt(2)sin2x+1 le2sinx + sqrt(2)cosx rArr 2sqrt(2)sinx cosx-2sinx - sqrt(2)cosx+1 le0`
`rArr 2sinx(sqrt(2)cosx-1)-1(sqrt(2)cosx-1) le0 rArr (2sinx-1)(sqrt(2)cosx-1) le0`
Above inequality holds when:
Case-I : `sinx-1/2 le0` and `cosx-1/sqrt(2)ge0 rArr sinx le1/2` and `cosxge1/sqrt(2)`
for `sinx le1/2: x in [-pi/2, pi/6] cup [(5pi)/6, (3pi)/2]` and for `cosx ge1/sqrt(2) : x in [-pi/4, pi/4]`
For both to simultaneously hold true: `x in [-pi/4, pi/6]`
Case-II `sinx-1/2 ge0` and `cosx le1/sqrt(2)`
Again, for the given interval of `x`:
For `sinx ge1/2 : x in [pi/5, (5pi)/6]` and for `cosx le1/sqrt(2): x in [-pi/2, -pi/4] cup [pi/4, (3pi)/2]`
For both to simultaneously hold true: `x in [pi/4, (5pi)/6]`
`therefore` Given inequality holds for `x in [-pi/4, pi/6] cup [pi/4, (5pi)/6]` Ans.