1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Can someone check my solution to a trigonometric equation?

  1. Feb 6, 2015 #1
    1. The problem statement, all variables and given/known data
    4sec2θ tanθ = 8 tanθ

    2. Relevant equations
    sec2θ = tan2θ + 1

    3. 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
     
  2. jcsd
  3. Feb 6, 2015 #2
    This looks completely correct to me
     
  4. Feb 6, 2015 #3

    Mark44

    Staff: Mentor

    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.
    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.
     
  5. Feb 6, 2015 #4
    Isn't it just θ = 3π/4 + kπ or θ = π/4 + kπ?

    Okay, that's useful to know.
     
  6. Feb 6, 2015 #5

    Mark44

    Staff: Mentor

    Yes, but since you didn't include any information about the expected solutions, I didn't know exactly what the problem was asking for.
     
  7. Feb 6, 2015 #6
    Oh, sorry.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted