How do you solve #cot^2x-cscx=1# and find all solutions in the interval #0<=x<360#?

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Answer 1 · 2016-10-17T13:20:23

#cos^2x/sin^2x - 1/sinx = 1#

#cos^2x/sin^2x -sinx/sin^2x - sin^2x/sin^2x = 0#

#cos^2x -sinx - sin^2x = 0#

Apply the identity #cos^2x = 1 - sin^2x#.

#1 - sin^2x - sinx - sin^2x = 0#

#-2sin^2x - sinx + 1 = 0#

#-2sin^2x - 2sinx + sinx + 1 = 0#

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

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

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

#x = 210˚, 330˚, 270˚#

Hopefully this helps!