Question

Find the values of x in the interval `[0,2pi]` which satisfy the inequality:
`3|2sinx-1|ge3+4cos^(2)x`

Answer 1

The given inequality can be written as:
`3|2sinx-1| ge3 +4(1-sin^(2)x) rArr 3|2sinx-1| ge7-4sin^(2)x`
Let `sinx=t rArr |3/2t -1|ge7-4t^(2)`
Case I: For `2t-1 ge0` i.e., `t ge 1//2` we have,` |2t-1|=(2t-1)`
`rArr 3(2t-1) ge 7-4t^(2) rArr 6t-3 ge 7-4t^(2)`
`rArr 4t^(2)+6y-10 ge0 rArr 2t^(2)+3t-5 ge0`
`rArr (t-1)(2t+5) ge0 rArr t le -5/2` and `t ge1`
Now for `t ge1/2`, we get `t ge1` from above conditions i.e., `sinx ge1`
The inequality holds true for x satisfying the equation `sinx=1 therefore x=pi/2`
(for `x in [0, 2pi])`
Case II: For `2t-t lt 0 rArr t lt 1/2`
We have, `|2t-1 lt 0| rArr t lt 1/2`
We have, `|2t-1|=-(2t-1)`
`rArr -3(2t-1) ge 7-4t^(2) rArr -6t+3 ge7-4t^(2)`
`rArr 4t^(2)-6t-4ge0 rArr 2t^(2)-3t-2 ge0`
`rArr (t-2)(2t+1) ge0 rArr t le -1/2` and `t ge2`
Again for `t lt 1/2`, we get `t le -1/2` from above conditions
i.e., `sinx le 1/2 rArr (7pi)/6 le x le 11/6 pi` (for `x in [0,2pi])`
Thus, `x in [(7pi)/6, (11pi)/6] cup {pi/2}` Ans.