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
Find all values of x, 0<x<(pi/2) for which (1/( root 3))<cotx<(root 3)
1 answer