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

- Babynini

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

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- Babynini

@peachpi the victory was short lived lol

- Babynini

3(theta)=arctan(-sq3/3)
?

- anonymous

This is the original?
\[\sqrt{3\tan 3\theta+1}=0\]

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## More answers

- Babynini

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

- anonymous

got you

- Babynini

\[\tan3(\theta)=\frac{ -\sqrt{3} }{ 3 }\]

- Babynini

right?

- anonymous

yes

- Babynini

So then after quite a bit of simplifying
theta = -0.26 ?

- Babynini

oh wait
-0.17

- 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

pi/18 is on the unit circle? o.o

- anonymous

Well pi/6 is.

- anonymous

You had 3Θ = arctan (-1/√3)
3Θ = -π/6
Θ = -π/18

- Babynini

Why does 3(theta) = -pi/6 ?

- Babynini

sorry, these are probably really elementary questions but I want to make sure and understand.

- 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

hm..k.

- anonymous

does that make sense?

- Babynini

Yeah I just hadn't made the connection so now i've got to think it through a bit :P

- Babynini

So now to find it in the right quadrants we add pi?

- 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

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

|dw:1434609190233:dw|

- anonymous

Dividing each of those by 3 gives
Θ = 11π/18 and Θ = 5π/18

- anonymous

Make sense so far?

- Babynini

and those aren't negative.

- anonymous

right. we don't want the negative solutions

- 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

so it would be writen as
5pi/18 + 2kpi ?

- anonymous

no just 5pi/18. because they gave a specific interval, you don't use 2kpi

- Babynini

ou.

- Babynini

but we're missing two more quadrants still, right?

- 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

....17pi/18 ?

- anonymous

yes. that's a solution
|dw:1434609720505:dw|

- Babynini

23pi/18 ?

- anonymous

yes

- Babynini

so our four answers are
5pi/18, 11pi/18. 17pi/18, 23pi/18

- anonymous

you've got 2 more to go

- Babynini

29pi/18?

- Babynini

35pi/18

- anonymous

you got it :)

- Babynini

yaya thaanks.

- anonymous

no problem

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