# Trig

How do I solve the following?

sec 3 theta/2 = 2 on the interval 0<theta<2 pi?

I tried to look in my textbook but it doesn't have an example. I also tried google and I was unsuccessful> How do I do this question? I know that the period for sec is 2 pi but I don't know how to figure out the interval it is positive 2 on. Any help is appreciated. Thank you.

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1. if sec 3Ø/2 = 2 then
cos 3Ø/2 = 1/2

I know cos π/3 = 1/2
so
3Ø/2 = π/3 or 3Ø/2 = 2π - π/3 = 5π/3
then
3Ø = 2π/3 or 3Ø = 10π/3
Ø = 2π/9 or Ø = 10π/9 , both within the interval given

The period of the original is 2π/(3/2) = 4π/3
so another value of Ø = 2π/9 + 4π/3 = 14π/9

so Ø = 2π/3, 12π/9, and 14π/9

(120, 240, and 280 degrees, all 3 work in the original equation and are between 0 and 360 degrees)

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