How do you solve #-8\sin \theta = 4\sqrt { 3}#?

1 Answer
Nghi N
Apr 24, 2017

#(4pi)/3 + 2kpi#
#(5pi)/3 + 2kpi#

Explanation:

#- 8sin t = 4sqrt3#
#sin t = -(4sqrt3)/8 = - sqrt3/2#
Trig table give 2 solutions:
a. #t = - pi/3 + 2kpi#, or #t = (5pi)/3 + 2kpi# (co-terminal)
b. #t = pi - (-pi/3) = pi + pi/3 = (4pi)/3 + 2kpi#

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