How do you solve the equation #sqrt3sectheta+2=0# for #-pi<=theta<=3pi#?

1 Answer
Nghi N
Dec 21, 2016

#(2pi)/3, (4pi)/3, (8pi)/3, (10pi)/3#

Explanation:

#sqrt3.sec t + 2 = 0#.
#sqrt3/(cos t) = - 2#
#sqrt3 = - 2cos t#
#cos t = - sqrt3/2#

Trig table and unit circle -->
a. For interval #(-pi, pi)# --> #cos t = -sqrt3/2# --> arc #t = +- (2pi)/3#
--> 2 solution arcs --># t = (2pi)/3# and #t = (4pi)/3# (co-terminal)
b. For interval #(-pi, pi + 2pi)# or interval #(-pi, 3pi)#,
add 2 solution arcs -->
#t = (2pi)/3 + 2pi = (8pi)/3# and #t = (4pi)/3 + 2pi = (10pi)/3#

Answers for #(-pi, 3pi)#:
#(2pi)/3, (4pi)/3, (8pi)/3, (10pi)/3)#

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