How do you solve #cot^2theta=sqrt3 cottheta# in the domain #0<=theta<=360#?

1 Answer
Nghi N.
Jun 23, 2018

#pi/6; (pi/2); (7pi)/6; (3pi)/2#

Explanation:

#cot^2 t - sqrt3cot t = 0#
#cot t(cot t - sqrt3) = 0#
Either factor should be zero.
a. cot t = 0
Unit circle gives -->
#t = pi/2# and #t = (3pi)/2#
b. #cot t = sqrt3#
Trig table and unit circle give -->
#t = pi/6# and #t = pi/6 + pi = (7pi)/6#

Related questions
Impact of this question
1364 views around the world
You can reuse this answer
Creative Commons License