Question

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

Answer 1

@peachpi the victory was short lived lol

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

got you

Answer 6

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

Answer 7

right?

Answer 8

yes

Answer 9

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

Answer 10

oh wait
-0.17

Answer 11

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

Answer 12

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

Answer 13

Well pi/6 is.

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

hm..k.

Answer 19

does that make sense?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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)