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anonymous

  • one year ago

Thanks for helping! I was given a pre calc problem which I simplified to: (tanx-2)(tanx-1)=0 The question asks me to solve from there. However, it doesn't ask me to solve in terms of 0 to 2pi, it asks for all solutions. I know that I have to add an extension once I find the final solution, but am having a hard time figuring out how to do so. I know it looks something like+n*pi*k or something like that. Thank You!

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  1. anonymous
    • one year ago
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    @Kainui @jim_thompson5910 @amistre64 @ganeshie8 @zepdrix @campbell_st @Loser66

  2. anonymous
    • one year ago
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    @Luigi0210 @robtobey @sammixboo @Zarkon @pooja195 @radar

  3. zepdrix
    • one year ago
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    Hmm the 2 is going to cause a bit of a problem :) Not a nice friendly angle there. But the 1 is ok.

  4. anonymous
    • one year ago
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    I think the one is pi/4?

  5. zepdrix
    • one year ago
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    Applying our Zero-Factor Property,\[\Large\rm \tan x-1=0\]Add 1 to each side,\[\Large\rm \tan x=1\] Ok good, yes this corresponds to pi/4 in the first quadrant.

  6. anonymous
    • one year ago
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    since it is for all solutions, should I add the extension? Or is there another thing I need to do?

  7. zepdrix
    • one year ago
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    \[\Large\rm x=\frac{\pi}{4}+k \pi, \quad k=0,\pm1,\pm2,...\]Tangent is a little different than sine and cosine. It is period in `pi` not in `2pi`. So we want to allow multiples of pi to be added on and give us the same value through tangent.

  8. zepdrix
    • one year ago
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    periodic* blah

  9. anonymous
    • one year ago
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    So I'd just say pi/4+kpi and add nEz to the right?

  10. zepdrix
    • one year ago
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    \(\Large\rm k\in\mathbb{Z}\) ya that works out nicely I guess :)

  11. anonymous
    • one year ago
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    cool and I don't need to add 5pi/4+kpi? As it is another solution

  12. zepdrix
    • one year ago
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    5pi/4 = pi/4 + 1*pi so it's already included in our set of solutions with the k

  13. zepdrix
    • one year ago
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    we just need to deal with this ugly tanx-2 now :d

  14. anonymous
    • one year ago
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    lol, I simplified it to 7pi/20

  15. anonymous
    • one year ago
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    I think that sounds right, but am not sure

  16. zepdrix
    • one year ago
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    Woah I'm not really sure 0_O I was just going to be lazy and call the angle \(\Large\rm \arctan(2)\)

  17. anonymous
    • one year ago
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    I asked my teacher, but he said I needed to go further

  18. zepdrix
    • one year ago
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    mmm ok thinking :)

  19. anonymous
    • one year ago
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    cool

  20. anonymous
    • one year ago
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    I found the radian in decimals and divided by pi and converted that into a fraction

  21. anonymous
    • one year ago
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    If that is the correct process

  22. anonymous
    • one year ago
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    and multiplied pi of course :)

  23. zepdrix
    • one year ago
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    arctan(2) = 1.1071487... divide by pi gives: \(\Large\rm 0.352416382\pi\) then what did you do? 0_o how did you convert to fraction?

  24. anonymous
    • one year ago
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    yea

  25. anonymous
    • one year ago
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    about 7/20

  26. anonymous
    • one year ago
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    is that right? I'm not too good at this ha ha

  27. anonymous
    • one year ago
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    I converted with some calculator

  28. anonymous
    • one year ago
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    which gave me 7/20, which is about right

  29. anonymous
    • one year ago
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    I rounded it for the calc

  30. zepdrix
    • one year ago
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    im not sure how to get an exact answer :d its look like it's some irrational multiple of pi, like some weird square root maybe. so ya approximating is probably the right way to go. I would leave it as a decimal though maybe. \[\Large\rm x=0.352\pi+k \pi\]\[\Large\rm x=(0.352+k)\pi,\qquad k\in\mathbb{Z}\]

  31. zepdrix
    • one year ago
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    Maybe you guys have some fun technique for finding arctan(2), maybe my brain is a lil rusty hehe

  32. anonymous
    • one year ago
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    thanks for the help! I'll give you a medal

  33. zepdrix
    • one year ago
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    np c:

  34. anonymous
    • one year ago
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    :)

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