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!

Solve for 0 ≤ θ ≤ 2π (pi)

  1. Apr 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve for 0 ≤ θ ≤ 2π (pi). Give the exact values.


    I'm not sure how to go about this one =/
  2. jcsd
  3. Apr 12, 2009 #2
    First off, observe that this equation holds when [tex]tan(\theta) = 0[/tex], so you should be able to find three solutions immediately.

    Now, try and find more solutions. Hint: you can safely divide by [tex]tan(\theta)[/tex] now because you know it's not equal to 0. Then, remember that
    [tex]tan(\theta) = \frac{sin(\theta)}{cos(\theta)}[/tex].

    Let's see what you can do from there!
  4. Apr 12, 2009 #3
    ok I worked it out and i got: cosθ=0 and cosθ=1/2 which gave me π/2 and π/3 ... does that seem right?
  5. Apr 12, 2009 #4
    The [tex]cos(\theta) = 1/2[/tex] is correct, but I don't know how you arrived at [tex]cos(\theta) = 0[/tex]. If [tex]cos(\theta) = 0[/tex], then [tex]tan(\theta)[/tex] is undefined. Although both sides are undefined, it does not mean they are equal. That would be similar to saying 1/0 = 0/0, which is not true! Also, remember there are two solutions for [tex]cos(\theta) = 1/2[/tex] on [tex][0,2\pi][/tex], one of them is [tex]\pi/3[/tex] as you mentioned. What do you think the other one is? Drawing a graph out might help.
  6. Apr 12, 2009 #5
    the other one would be 5π/3

    here's what I did

    sin^2 θ/cos^2 θ=2(sinθ/cosθ).sinθ
    sin^2 θ/cos^2 θ=2(sin^2 θ/cosθ)
    1/cos^2 θ=2/cosθ
    2cos^2 θ-cosθ=0
    cosθ(2cosθ -1)=0

    cosθ=0 and 2cosθ -1=0
  7. Apr 12, 2009 #6
    Ok, I see what you did now. The reason that doesn't work is because, as I mentioned in my previous post, [tex]tan(\theta)[/tex] is undefined when [tex]cos(\theta) = 0[/tex]. So, this is not a valid answer.

    Here's what I was attempting to get you to do:

    [tex]tan^{2}(\theta) = 2tan(\theta)sin(\theta)
    => tan(\theta) = 2sin(\theta)
    => \frac{sin(\theta)}{cos(\theta)} = 2sin(\theta)
    => \frac{1}{cos(\theta)} = 2
    => cos(\theta) = \frac{1}{2}

    You can divide by [tex]tan(\theta)[/tex] and [tex]sin(\theta)[/tex] because we assume that [tex]tan(\theta) \ne 0[/tex] because we already found the solutions when [tex]tan(\theta) = 0[/tex].

    Is this clear?
  8. Apr 12, 2009 #7
    Generally if you are stuck, put everything in terms of cosines and sines so that you can combine and cancel things.

    [tex]\cos{\theta}[/tex] doesn't work because tangent is undefines when cosθ=0. You also divided by sinθ, which assumes that [tex]\sin{\theta}\not= 0[/tex]. This caused you to lose that solution which works.
  9. Apr 12, 2009 #8
    ah ok thanks a ton guys =]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Solve for 0 ≤ θ ≤ 2π (pi)
  1. Cot^(-1)(0) = pi/2, why? (Replies: 10)