Correct Answer - B
We observe that `"log"_("cos"x)sin x` is defined for ` x in (0, pi//2) cup (2pi, 5pi//2)`.
Now,
`"log_("cos"x) "sin"x ge 2`
`rArr "sin" x le ("cos" x)^(2) " "[because 0 lt "cos" x lt 1]`
`rArr "sin"^(2) x + "sin"x -1 lt 0`
`rArr ("sin" x + (1)/(2))^(2) - (5)/(4) le 0`
`rArr ("sin" x + (1)/(2) - (sqrt(5))/(2))("sin" x + (1)/(2) + (sqrt(5))/(2)) le 0`
`rArr "sin" x + (1)/(2) -(sqrt(5))/(2) le 0 " " [because "sin" x + (1)/(2) + (sqrt(5))/(2) gt 0]`
`rArr "sin" x le (sqrt(5)-1)/(2) rArr "sin" x in (0, (sqrt(5)-1)/(2)]`
