Asked by Jeremy
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
17sec2(θ)-13tan(θ)sec(θ)-15=0
Answers
Answered by
plumpycat
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°.
Answered by
Steve
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
...
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
...
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.