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Abdallah2012

  • one year ago

tan theta + sec theta = 1 between 0 and 2 pi

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  1. swissgirl
    • one year ago
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    can theta=0 ? Like is is\[ 0< \theta <\Pi\] Or \[ 0 \leq \theta \leq \Pi\] ?

  2. Abdallah2012
    • one year ago
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    0<θ<2Π has to be between this interval

  3. RED123url
    • one year ago
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    use substitution y=tan(theta/2) |dw:1383969602615:dw|

  4. Abdallah2012
    • one year ago
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    0≤θ<2Π sorry correction, it has to be between this interval

  5. RED123url
    • one year ago
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    with a little bit of work you will find the solution

  6. Karina1
    • one year ago
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    @Mertsj what about if they have a log like for example using the graph solve the system y=-2x+3 and y=logx+1 im confused cause of the log

  7. Abdallah2012
    • one year ago
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    @RED123url I am not sure what you did there...

  8. RED123url
    • one year ago
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    well y=tan(theta/2) sin(thata)= 2y/(y^2+1) cos(theta)=(1-y^2)/(y^2+1) you know tan=sin/cos sec=1/cos plug things in and you will get an equation of y then go back y=tan)theta/2) and use the inverse

  9. swissgirl
    • one year ago
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    The answer is 0 but I used trial and error

  10. Abdallah2012
    • one year ago
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    @RED123url so just to confirm what swissgirl got, did you get the same answer?

  11. swissgirl
    • one year ago
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    I graphed it to confirm my answer and its def right

  12. RED123url
    • one year ago
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    it should be 2pi

  13. swissgirl
    • one year ago
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    Well it is 0 and 2 pi But if you noticed the interval is \[ 0 \leq \theta < 2\pi \]

  14. Abdallah2012
    • one year ago
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    @RED123url well 2pi is not one the answers but 0 is, also according to the interval 2pi cannot be the answer..

  15. RED123url
    • one year ago
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    the real answer for these kind of problem involves n: the answer is 2npi. n can be 0 or any number even a billion

  16. Abdallah2012
    • one year ago
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    @RED123url so would you agree that in this case the answer is 0?

  17. RED123url
    • one year ago
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    if 2pi is excluded of course

  18. Abdallah2012
    • one year ago
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    well then. I am going with 0. Thank you both for your help!

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