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Solve for 0 ≤ θ ≤ 2π (pi)

  • Thread starter rought
  • Start date
  • #1
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Homework Statement



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

tan^2θ=2tanθsinθ


I'm not sure how to go about this one =/
 

Answers and Replies

  • #2
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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!
 
  • #3
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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?
 
  • #4
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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.
 
  • #5
34
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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.
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
 
  • #6
56
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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?
 
  • #7
290
2
I'm not sure how to go about this one =/
Generally if you are stuck, put everything in terms of cosines and sines so that you can combine and cancel things.

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
[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.
 
  • #8
34
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ah ok thanks a ton guys =]
 

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