Question

Hi guys, I was wondering if you guys can help me with this question. I'm not sure how to do this question. As I keep on getting something that's out of the unit circle. Please help!
17sec2(θ)-13tan(θ)sec(θ)-15=0

Answer 1

17/cos² θ - 13(sin θ/cos θ)(1/cos θ) - 15 = 0
17/cos² θ - 13(sin θ/cos² θ) - 15 = 0
17 - 13sin θ -15cos² θ = 0
17 - 13sin θ -15(1-sin ² θ) = 0
15sin ² θ - 13sin θ + 2 = 0

This will now factor nicely across the integers, to give you two solutions within 0 ≤ θ ≤ 360°.

Answer 2

17sec^2(θ)-13tan(θ)sec(θ)-15=0
17+17tan^2θ - 13secθtanθ - 15 = 0
13 secθtanθ = 17tan^2θ + 2
169 sec^2θ tan^2θ = 289tan^4θ + 68tan^2θ + 4
169tan^4θ + 169tan^2θ = 289tan^4θ + 68tan^2θ + 4
120tan^4θ - 101tan^2θ + 4 = 0
(24tan^2θ-1)(5tan^2θ-4) = 0

tan^2θ = 1/24 or 4/5

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