How do you solve #tan(2x + pi/3) = 1# from 0 to 2pi?

1 Answer
Nov 6, 2015

#(11pi)/24 ; (23pi)/24 ; (33pi)/24 , (35pi)/24#

Explanation:

Trig Table and unit circle -->
# tan (pi/4) = 1# and #tan (pi/4 + pi) = tan ((5pi)/4) = 1#
Substitute in the right side of the equation 1 by tan (pi/4), and 1 by tan ((5pi)/4):

a. #tan (2x + pi/3) = 1 = tan (pi/4)#
#2x + pi/3 = pi/4 + kpi#
#2x = pi/4 - pi/3 = - pi/12 + kpi# --> #x1 = - pi/24 + (kpi)/2#
#x1 = - pi/24 = ((23pi)/24) + (kpi)/2# (co-terminal arc)
If k = 1 --> #x2 = (23pi)/24 + pi/2 = (35pi)/24#
b. #tan (2x + pi/3) = tan ((5pi)/4)#
#2x + pi/3 = (5pi)/4 + kpi#
#2x = (5pi)/4 - pi/3 = (11pi)/12 + kpi# --> #x3 = (11pi)/24 + (kpi)/2#
If k = 1 --> #x4 = (11pi)/24 + pi/2 = (33pi)/24#
Answers for #(0, 2pi)#:
#(11pi)/24 , (23pi)/24 , (33pi)/24, (35pi)/24#