How do you solve #\tan \theta = - 2\sin \theta#?

1 Answer
Nghi N
Feb 23, 2017

#t = kpi#
#t = +- (2pi)/3 + 2kpi#

Explanation:

sin t/(cos t) = - 2sin t
sin t = - 2sint.cos t
sin t(1 + 2cos t) = 0
a. sin t = 0 --> unit circle --> #t = 0, t = pi, t = 2pi#
General answers: #t = kpi#
b. 1 + cos t = 0 --> # cos t = - 1/2#
Trig table and unit circle -->
#t = +- (2pi)/3 + 2kpi#,

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