How do you solve #2= - 2+ 3\sec ^ { 2} \theta#?

1 Answer
Nghi N
Jun 21, 2017

#t = +- pi/6 + 2kpi#
#t = +- (5pi)/6 + 2kpi#

Explanation:

#2 = -2 + 3sec^2 t#
#sec^2 t = 4/3#
#sec t = +- 2/(sqrt3)#
#cos t = 1/(sec t) = +- sqrt3/2#
Trig table and unit circle give:

a. #cos t = sqrt3/2# -->
#t = +- pi/6 + 2kpi#
b. #cos t = - sqrt3/2# -->
#t = +- (5pi)/6 + 2kpi#

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