How do you solve #cot^2xcsc^2x-cot^2x=9# for #0<=x<=2pi#?

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Answer 1 · 2016-08-08T09:33:20

# x in {pi/3, 2pi/3, 4pi/3, 5pi/3} sub [0,2pi]#.

#cot^2xcsc^2x-cot^2x=9#

#:. (csc^2x-1)cot^2x=9#

#:.cot^2x*cot^2x=9#

#:. cot^4x=9#

#:. cotx=+-sqrt3#

#:. tanx=+-1/sqrt3=tan(+-pi/6)#

#:. x in {npi+-pi/3 ; n in ZZ}#

#:. x in {pi/3, 2pi/3, 4pi/3, 5pi/3} sub [0,2pi]#.