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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.

    tan^2θ=2tanθsinθ


    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

    tan^2θ=2tanθsinθ
    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}
    [/tex]

    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 =]
     
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