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Babynini

  • one year ago

Find all solutions for equation, find solutions in interval [0,2pi) sq3 tan 3(theta)+1=0

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  1. Babynini
    • one year ago
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    @peachpi the victory was short lived lol

  2. Babynini
    • one year ago
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    3(theta)=arctan(-sq3/3) ?

  3. anonymous
    • one year ago
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    This is the original? \[\sqrt{3\tan 3\theta+1}=0\]

  4. Babynini
    • one year ago
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    hm, no. Let me write it out for you :) \[\sqrt{3}\tan3(\theta)+1=0\]

  5. anonymous
    • one year ago
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    got you

  6. Babynini
    • one year ago
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    \[\tan3(\theta)=\frac{ -\sqrt{3} }{ 3 }\]

  7. Babynini
    • one year ago
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    right?

  8. anonymous
    • one year ago
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    yes

  9. Babynini
    • one year ago
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    So then after quite a bit of simplifying theta = -0.26 ?

  10. Babynini
    • one year ago
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    oh wait -0.17

  11. anonymous
    • one year ago
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    That one's on the unit circle, so they might be looking for an exact answer. -π/18, which is about -0.17, so you're right

  12. Babynini
    • one year ago
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    pi/18 is on the unit circle? o.o

  13. anonymous
    • one year ago
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    Well pi/6 is.

  14. anonymous
    • one year ago
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    You had 3Θ = arctan (-1/√3) 3Θ = -π/6 Θ = -π/18

  15. Babynini
    • one year ago
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    Why does 3(theta) = -pi/6 ?

  16. Babynini
    • one year ago
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    sorry, these are probably really elementary questions but I want to make sure and understand.

  17. anonymous
    • one year ago
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    because arctan has a domain from -pi/2 to pi/2 (because tan positive in the 1st quadrant, negative in the 4th). The angle in that domain with a tan of -1/√3 is -π/6

  18. Babynini
    • one year ago
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    hm..k.

  19. anonymous
    • one year ago
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    does that make sense?

  20. Babynini
    • one year ago
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    Yeah I just hadn't made the connection so now i've got to think it through a bit :P

  21. Babynini
    • one year ago
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    So now to find it in the right quadrants we add pi?

  22. anonymous
    • one year ago
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    ok so this one's a little trickier since the tan is negative. The restriction on the equation is 0 to 2pi, so we actually need to add 2pi to -pi/6 to get the 1st solution for 3pi

  23. anonymous
    • one year ago
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    We'll have 2 solutions for 3pi between 0 and 2pi: 11π/6 in the 4th quadrant (-pi/6 + 2π) 5π/6 in the 2nd quadrant (-π/6 + π) |dw:1434608981061:dw|

  24. anonymous
    • one year ago
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    |dw:1434609190233:dw|

  25. anonymous
    • one year ago
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    Dividing each of those by 3 gives Θ = 11π/18 and Θ = 5π/18

  26. anonymous
    • one year ago
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    Make sense so far?

  27. Babynini
    • one year ago
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    and those aren't negative.

  28. anonymous
    • one year ago
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    right. we don't want the negative solutions

  29. anonymous
    • one year ago
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    You need to find 3Θ for 0<Θ<6π, so add 2pi to both 5π/6 and 11π/6 twice to get the other 4 solutions for 3Θ

  30. Babynini
    • one year ago
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    so it would be writen as 5pi/18 + 2kpi ?

  31. anonymous
    • one year ago
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    no just 5pi/18. because they gave a specific interval, you don't use 2kpi

  32. Babynini
    • one year ago
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    ou.

  33. Babynini
    • one year ago
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    but we're missing two more quadrants still, right?

  34. anonymous
    • one year ago
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    there are 4 more solutions. You need to find 3Θ for 0<Θ<6π, so add 2pi to both 5π/6 and 11π/6 twice to get the other 4 solutions for 3Θ

  35. Babynini
    • one year ago
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    ....17pi/18 ?

  36. anonymous
    • one year ago
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    yes. that's a solution |dw:1434609720505:dw|

  37. Babynini
    • one year ago
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    23pi/18 ?

  38. anonymous
    • one year ago
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    yes

  39. Babynini
    • one year ago
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    so our four answers are 5pi/18, 11pi/18. 17pi/18, 23pi/18

  40. anonymous
    • one year ago
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    you've got 2 more to go

  41. Babynini
    • one year ago
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    29pi/18?

  42. Babynini
    • one year ago
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    35pi/18

  43. anonymous
    • one year ago
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    you got it :)

  44. Babynini
    • one year ago
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    yaya thaanks.

  45. anonymous
    • one year ago
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    no problem

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