Posted by Patrick on Wednesday, October 10, 2012 at 7:22pm.
0<x<(π/2) ---> x must be in quadrant I
1/√3 < cotx < √3
consider the "ends"
cotx = 1/√3 ---> tanx = √3 ---> x = π/3 or (60°)
cotx = √3 ---> tanx = 1/√3 ---> x = π/6 or (30°)
so π/6 < x < π/3
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