How do I solve the equation #cot({5pi}/12-x)+1=0#?

1 Answer
Jun 8, 2018

#x = {2pi}/3 + pi k quad # integer #k#

Explanation:

#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#