# Trig

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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

• Trig -

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.

• Trig -

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°

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