How do you solve #cot^2x-cscx=1#?

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Answer 1 · 2017-01-03T15:32:51

#x = pi/6 + 2pin, (5pi)/6 + 2pin and (3pi)/2 + 2pin#

Use the pythagorean identity #cot^2theta + 1 = csc^2theta#. This means that #cot^2theta = csc^2theta - 1#.

#csc^2x - 1 - cscx = 1#

#csc^2x - cscx - 2 = 0#

Let #t = cscx#.

#t^2 - t - 2 = 0#

#(t - 2)(t + 1) = 0#

#t = 2 and -1#

#cscx = 2 and cscx = -1#

#1/sinx = 2 and 1/sinx = -1#

#sinx = 1/2 and sinx = -1#

#x = pi/6 + 2pin, (5pi)/6 + 2pin and (3pi)/2 + 2pin#

Hopefully this helps!