Asked by Lisa
Find all solutions in the interval 0 degrees<θ<360 degrees. If rounding necessary, round to the nearest tenth of a degree. 17sec2 θ − 15tanθsecθ − 15 = 0
Answers
Answered by
Steve
17sec^2θ - 15secθ tanθ - 15 = 0
15secθ tanθ = 17sec^2θ - 15
225 sec^2θ tan^2θ = 289sec^4θ - 510sec^2θ + 225
225sec^4θ - 225sec^2θ = 289sec^4θ - 510sec^2θ + 225
64sec^4θ - 285sec^2θ + 225 = 0
That's just a quadratic in sec^2θ, so just solve it and you have your solution candidates.
However, because we squared things, there may be spurious solutions, so you have to check the values in the original equation.
In radians, I get .159,1.000,2.141,2.982
15secθ tanθ = 17sec^2θ - 15
225 sec^2θ tan^2θ = 289sec^4θ - 510sec^2θ + 225
225sec^4θ - 225sec^2θ = 289sec^4θ - 510sec^2θ + 225
64sec^4θ - 285sec^2θ + 225 = 0
That's just a quadratic in sec^2θ, so just solve it and you have your solution candidates.
However, because we squared things, there may be spurious solutions, so you have to check the values in the original equation.
In radians, I get .159,1.000,2.141,2.982
Answered by
Reiny
17sec2 θ − 15tanθsecθ − 15 = 0
17/cos^2 Ø - 15(sinØ/cosØ)(1/cosØ) - 15 = 0
times cos^2 Ø
17 - 15sinØ - 15cos^2 Ø = 0
17 - 15sinØ - 15(1 - sin^2 Ø) = 0
15sin^2 Ø - 15sinØ + 2 = 0
sinØ = (15 ± √105)/30
sinØ = .158435 or sinØ = .841565
Ø = 9.1° or 170.9° or Ø = 57.3° or 122.7°
17/cos^2 Ø - 15(sinØ/cosØ)(1/cosØ) - 15 = 0
times cos^2 Ø
17 - 15sinØ - 15cos^2 Ø = 0
17 - 15sinØ - 15(1 - sin^2 Ø) = 0
15sin^2 Ø - 15sinØ + 2 = 0
sinØ = (15 ± √105)/30
sinØ = .158435 or sinØ = .841565
Ø = 9.1° or 170.9° or Ø = 57.3° or 122.7°
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