Babynini
  • Babynini
Find all solutions for equation, find solutions in interval [0,2pi) sq3 tan 3(theta)+1=0
Mathematics
chestercat
  • chestercat
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Babynini
  • Babynini
@peachpi the victory was short lived lol
Babynini
  • Babynini
3(theta)=arctan(-sq3/3) ?
anonymous
  • anonymous
This is the original? \[\sqrt{3\tan 3\theta+1}=0\]

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Babynini
  • Babynini
hm, no. Let me write it out for you :) \[\sqrt{3}\tan3(\theta)+1=0\]
anonymous
  • anonymous
got you
Babynini
  • Babynini
\[\tan3(\theta)=\frac{ -\sqrt{3} }{ 3 }\]
Babynini
  • Babynini
right?
anonymous
  • anonymous
yes
Babynini
  • Babynini
So then after quite a bit of simplifying theta = -0.26 ?
Babynini
  • Babynini
oh wait -0.17
anonymous
  • anonymous
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
Babynini
  • Babynini
pi/18 is on the unit circle? o.o
anonymous
  • anonymous
Well pi/6 is.
anonymous
  • anonymous
You had 3Θ = arctan (-1/√3) 3Θ = -π/6 Θ = -π/18
Babynini
  • Babynini
Why does 3(theta) = -pi/6 ?
Babynini
  • Babynini
sorry, these are probably really elementary questions but I want to make sure and understand.
anonymous
  • anonymous
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
Babynini
  • Babynini
hm..k.
anonymous
  • anonymous
does that make sense?
Babynini
  • Babynini
Yeah I just hadn't made the connection so now i've got to think it through a bit :P
Babynini
  • Babynini
So now to find it in the right quadrants we add pi?
anonymous
  • anonymous
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
anonymous
  • anonymous
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|
anonymous
  • anonymous
|dw:1434609190233:dw|
anonymous
  • anonymous
Dividing each of those by 3 gives Θ = 11π/18 and Θ = 5π/18
anonymous
  • anonymous
Make sense so far?
Babynini
  • Babynini
and those aren't negative.
anonymous
  • anonymous
right. we don't want the negative solutions
anonymous
  • anonymous
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
Babynini
  • Babynini
so it would be writen as 5pi/18 + 2kpi ?
anonymous
  • anonymous
no just 5pi/18. because they gave a specific interval, you don't use 2kpi
Babynini
  • Babynini
ou.
Babynini
  • Babynini
but we're missing two more quadrants still, right?
anonymous
  • anonymous
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
Babynini
  • Babynini
....17pi/18 ?
anonymous
  • anonymous
yes. that's a solution |dw:1434609720505:dw|
Babynini
  • Babynini
23pi/18 ?
anonymous
  • anonymous
yes
Babynini
  • Babynini
so our four answers are 5pi/18, 11pi/18. 17pi/18, 23pi/18
anonymous
  • anonymous
you've got 2 more to go
Babynini
  • Babynini
29pi/18?
Babynini
  • Babynini
35pi/18
anonymous
  • anonymous
you got it :)
Babynini
  • Babynini
yaya thaanks.
anonymous
  • anonymous
no problem

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