Wednesday

March 4, 2015

March 4, 2015

Posted by **Patrick** on Wednesday, October 10, 2012 at 7:22pm.

- Precalc/Trig -
**Reiny**, Wednesday, October 10, 2012 at 7:46pm0<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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