- #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 =/

- Thread starter rought
- Start date

- #1

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

tan^2θ=2tanθsinθ

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

- #2

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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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- #4

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- #5

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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

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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

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Generally if you are stuck, put everything in terms of cosines and sines so that you can combine and cancel things.I'm not sure how to go about this one =/

[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.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

- #8

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ah ok thanks a ton guys =]

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