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Dama28

  • 2 years ago

Solve the equation! 3 tan^3θ = tan θ

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  1. tkhunny
    • 2 years ago
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    Just like you used to solve polynomials back in algebra. Collecte everything to one side. Factor everything. Give it a go!

  2. cnknd
    • 2 years ago
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    lol dont need to go that far. let x = tan(theta), then 3x^3 = x, so either x = 0, or you can divide x from both sides to get: 3x^2 = 1 so x can be 0, 1/sqrt(3), or -1/sqrt(3).

  3. tkhunny
    • 2 years ago
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    In other words, collect and factor. Of course, you should solve the actual problem by finding the angle that gives these tangents.

  4. Dama28
    • 2 years ago
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    now I need to find all real solutions with k being any integer.

  5. tkhunny
    • 2 years ago
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    You have \(\tan(\theta) = 0\), \(\tan(\theta) = \dfrac{1}{\sqrt{3}}\), and \(\tan(\theta) = \dfrac{-1}{\sqrt{3}}\). Track them down.

  6. Dama28
    • 2 years ago
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    ok, so...for tan(θ)=0 I get 0 and pi. for tan(θ)=1/√3 I get pi/6 and 7pi/6. And for tan(θ)=−1/√3 I get 5pi/6 and 11pi/6. how do I know if I would add pi k to each or 2pi k?

  7. lopus
    • 2 years ago
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    |dw:1353819049038:dw|

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