Can someone check my solution to a trigonometric equation?

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Homework Help Overview

The discussion revolves around solving a trigonometric equation involving secant and tangent functions. The original poster presents their solution process for the equation 4sec²θ tanθ = 8 tanθ, using the identity sec²θ = tan²θ + 1.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the steps taken to simplify the equation and solve for θ. There is a focus on the correctness of the solutions provided, particularly regarding the range of solutions and the notation used.

Discussion Status

Some participants express confidence in the original poster's solution, while others raise concerns about the completeness of the work in relation to the problem's requirements. Clarifications regarding the notation and the expected range of solutions are being explored.

Contextual Notes

There is uncertainty about whether the problem is asking for solutions within a specific interval or for all possible solutions. This has led to differing interpretations of the original poster's work.

Eclair_de_XII
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Homework Statement


4sec2θ tanθ = 8 tanθ

Homework Equations


sec2θ = tan2θ + 1

The Attempt at a Solution


4(tan2 + 1) tanθ = 8 tanθ
(4tan2 + 4) tanθ = 8 tanθ
4 tan3 + 4 tanθ = 8 tanθ
4 tan3 - 4 tanθ = 0
4tanθ(tan2θ - 1) = 0

4tanθ = 0
tanθ = 0
tan-1(tanθ) = tan-1(0)
tan-1θ = 0, π

tan2θ - 1 = 0
(tanθ + 1)(tanθ - 1) = 0

tanθ + 1 = 0
tanθ = -1
tan-1tanθ = tan-1(-1)
tan-1θ = 3π/4, 7π/4

tanθ - 1 = 0
tanθ = 1
tan-1tanθ = tan-1(1)
tan-1θ = π/4, 5π/4
 
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This looks completely correct to me
 
Eclair_de_XII said:

Homework Statement


4sec2θ tanθ = 8 tanθ

Homework Equations


sec2θ = tan2θ + 1

The Attempt at a Solution


4(tan2 + 1) tanθ = 8 tanθ
(4tan2 + 4) tanθ = 8 tanθ
4 tan3 + 4 tanθ = 8 tanθ
4 tan3 - 4 tanθ = 0
4tanθ(tan2θ - 1) = 0

4tanθ = 0
tanθ = 0
tan-1(tanθ) = tan-1(0)
tan-1θ = 0, π

tan2θ - 1 = 0
(tanθ + 1)(tanθ - 1) = 0

tanθ + 1 = 0
tanθ = -1
tan-1tanθ = tan-1(-1)
tan-1θ = 3π/4, 7π/4

tanθ - 1 = 0
tanθ = 1
tan-1tanθ = tan-1(1)
tan-1θ = π/4, 5π/4
Your solutions above are fine if the question is asking for solutions θ such that 0 ≤ θ ≤ 2π, but if the question is asking for all solutions, then the work above is not complete.

I should also add that some of what you wrote is incorrect.
tan-1θ = 0, π
This should be θ = 0 or θ = π -- the tan-1 business shouldn't be there. The same is true on the other two sections where you have done this.
 
Mark44 said:
but if the question is asking for all solutions, then the work above is not complete.

Isn't it just θ = 3π/4 + kπ or θ = π/4 + kπ?

Mark44 said:
This should be θ = 0 or θ = π -- the tan-1 business shouldn't be there. The same is true on the other two sections where you have done this.

Okay, that's useful to know.
 
Eclair_de_XII said:
Isn't it just θ = 3π/4 + kπ or θ = π/4 + kπ?
Yes, but since you didn't include any information about the expected solutions, I didn't know exactly what the problem was asking for.
 
Oh, sorry.
 

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