Solve : `2 cos^(2)theta + sin theta le 2`, where `pi/2 le theta le (3pi)/2.`

Explanation

`2 cos^(2)theta + sin theta le 2`
`rArr` `2(1 - sin^(2)theta) + sin theta le 2`
`rArr` `-2 sin^(2)theta + sin theta le 0`
`rArr` `2 sin^(2)theta - sin theta ge 0`
`rArr` `sintheta(2 sin theta -1) ge 0`
`rArr` `sintheta(sin theta -1//2) ge 0`,
which is possible if `sin theta le 0` or `sin theta ge 1//2`

Now `sin theta ge 1//2 rArr pi//2 le theta le 5pi//6` (from the figure)
And `sin theta le 0 rArr pi le theta le 3pi//2` (from the figure)
Hence the required values of `theta` are given by `theta in [pi//2, 5pi//6] uu [pi, 3pi//2]`