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

Answer

`x in [pi//2, 5pi//6] uu [pi, 3pi//2]`

Explanation

We have `2 cos^(2) x+sin x le 2`
`:. 2(1- sin^(2)x)+sin x le 2`
`rArr -2 sin^(2) x+sin x le 0`
`rArr 2 sin^(@) x- sin x ge 0`
`rArr sin x(2 sin x-1) ge 0`
`rArr sin x(sin x -1//2) ge 0`
`rArr sin x le 0 or sin x ge 1//2`
Draw the graph of `y= sin x` and `y=1//2`.

From the figure,
`sin x ge 1//2`
`rArr n//2 le x le 5pi//6`
and `sin x le 0`
`rArr pi le x le 3 pi//2`
Hence the required solution is `x in [pi//2, 5pi//6] uu [pi, 3 pi//2]`.