#cot({5pi}/12 - x) + 1 = 0#
#cot({5pi}/12 - x) = - 1#
Cotangents are reciprocal slopes, so a cotangent of #-1# is a tangent of #1/-1=-1# too so #-45^circ# or #135^circ# and their coterminal friends.
#cot({5pi}/12 - x) = cot ( - pi/4) #
#{5 pi}/12 - x = -pi/4 + pi k quad # integer #k#
#x = {5pi}/12 + pi/4 + pi k quad # integer #k#
(It's OK to flip the sign on #k#, which still ranges over the integers.)
#x = {2pi}/3 + pi k quad # integer #k#
Check: Let's just check #k=-1# so #x=-pi/3#
#cot({5pi}/12 - (-pi/3)) + 1 = cot( {3pi}/4) + 1 = -1 + 1 = 0 quad sqrt#
