How do you solve the equation sqrt3sectheta+2=0√3secθ+2=0 for -pi<=theta<=3pi−π≤θ≤3π?

1 Answer
Nghi N
Dec 21, 2016

(2pi)/3, (4pi)/3, (8pi)/3, (10pi)/32π3,4π3,8π3,10π3

Explanation:

sqrt3.sec t + 2 = 0√3.sect+2=0.
sqrt3/(cos t) = - 2√3cost=−2
sqrt3 = - 2cos t√3=−2cost
cos t = - sqrt3/2cost=−√32

Trig table and unit circle -->
a. For interval (-pi, pi)(−π,π) --> cos t = -sqrt3/2cost=−√32 --> arc t = +- (2pi)/3t=±2π3
--> 2 solution arcs --> t = (2pi)/3t=2π3 and t = (4pi)/3t=4π3 (co-terminal)
b. For interval (-pi, pi + 2pi)(−π,π+2π) or interval (-pi, 3pi)(−π,3π),
add 2 solution arcs -->
t = (2pi)/3 + 2pi = (8pi)/3t=2π3+2π=8π3 and t = (4pi)/3 + 2pi = (10pi)/3t=4π3+2π=10π3

Answers for (-pi, 3pi)(−π,3π):
(2pi)/3, (4pi)/3, (8pi)/3, (10pi)/3)2π3,4π3,8π3,10π3)

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