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

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

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

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

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

$:. cot^4x=9$

$:. cotx=+-sqrt3$

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

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

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

How do you solve cot^2xcsc^2x-cot^2x=9cot^2xcsc^2x-cot^2x=9 for 0<=x<=2pi0<=x<=2pi?