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Can someone check my solution to a trigonometric equation?

  • #1
804
37

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
 

Answers and Replies

  • #2
63
1
This looks completely correct to me
 
  • #3
33,262
4,963

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.
 
  • #4
804
37
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?

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.
 
  • #5
33,262
4,963
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.
 
  • #6
804
37
Oh, sorry.
 

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