Solving a trigonometric equation help

[meeklobraca]

Homework Statement

sqrt3 cot 3x + 1 = 0 where 0 _<x_<2pie

Homework Equations

The Attempt at a Solution

I narrowed it down to a constant on one side,

cot 3x = -1/sqrt3 which I could turn into cos x/sin x = -1/sqrt3

but I have no idea where to go from here.

Any help would be greatly appreciated!

 

[gabbagabbahey]

Hmmm... if \cot(3x)=\frac{-1}{\sqrt{3}}, then \tan(3x)=?

 

[meeklobraca]

well it would be 1/-1/sqrt3 wouldn't it?

 

[gabbagabbahey]

well it would be 1/-1/sqrt3 wouldn't it?

Sure, but \frac{1}{\frac{-1}{\sqrt{3}}}=-\sqrt{3} right?

Do you know what angle(s) has a tangent of -\sqrt{3}?

 

[meeklobraca]

I don't actually. I am only familiar with the unit circle for sin/cos.

 

[jgens]

As tan(x) is defined as sin(x)/cos(x) you should be able to find the values for tan(x).

 

[gabbagabbahey]

Try checking every 30 degrees (\pi/6 radians) between 0 and 180 degrees. You should find one value in that range.

Then you know that you want to find all values for 0\leq x\leq2\pi, so that means that 3x goes from 0 to 6pi Right?

That means you need to find all angles between zero and 6pi where the tangent of those angles is -sqrt{3}. You can use the fact that Tangent is periodic to do that by adding multiples of 180 degrees o the angle you found earlier.

 

[meeklobraca]

okay well one second here, to find values of tan x using sin x/cos x, do i calculate that with a calculator to find the angle or use the unit circle, cause I am looking at the unit circle and there are no values for sin or cos sqrt3.

 

[gabbagabbahey]

use the unit circle: there are values where \sin(\theta)=\frac{\sqrt{3}}{2} and \cos(\theta)=\frac{-1}{2} aren't there?

Follow my instructions from my previous post; what is \tan(0)? How about \tan(\pi/6)? How about \tan(\pi/3) ...etc.?

 

[meeklobraca]

I don't get how you get those sin and cos values from tan sqrt 3.

Like using a calculator I get a value of -60. Then adding 180 degrees I get -60, 120, 300, 480, 660, 840, and 1020 for 3x.

 

[gabbagabbahey]

Just plug in angles every 30 degrees from 0 to 180...you'll see where i get them from

 

[meeklobraca]

Whats the process of plugging in angles every 30 degrees for tan 3x = sqrt3?

 

[gabbagabbahey]

\tan(0)=\frac{\sin(0)}{\cos(0)}=\frac{0}{1}=0

\tan(30)=\frac{\sin(30)}{\cos(30)}=\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}

\tan(60)=?

\tan(90)=?

\tan(120)=?

\tan(150)=?

\tan(180)=?

 

[meeklobraca]

Okay I gotcha there.

So the angles I've found are correct then? 120, 300? I assume I don't include the -60 because our range is 0 to 2pie?

 

[gabbagabbahey]

Okay I gotcha there.

So the angles I've found are correct then? 120, 300? I assume I don't include the -60 because our range is 0 to 2pie?

If 3x=120, then x=40...you should end up with 6 values for x in between 0 and 2pi

 

[meeklobraca]

So with this then, I have my two angles of 2pie/3 and 5pie/3, is 3x 2pie/6 5pie/6 and 2pie/9 and 5pie/9? Are those my 6 angles?

 

[NoMoreExams]

OK just have to say this, pi not pie

 

[meeklobraca]

ahhaha, I am sorry i should know better.

 

[meeklobraca]

Is my answer is post #16 correct? Or on the right track at least?

 

[NoMoreExams]

The beauty of math at your level is that you can just plug your answer in and see if it's right.

 

[meeklobraca]

Has anybody ever told you, your not much help?

 

[gabbagabbahey]

He was suggesting that you check your answers yourself by plugging them into your original equation. It is a good habit to get into because then you don't need to wait however long it might take for someone else to check them for you.

 

[meeklobraca]

I wouldn't have the slightest idea how to go about that with an equation that involves cot. Plus I shouldn't have to check. I am trying to figure out how to do these questions, and I can't rely on double checking the answers after.

 

[gabbagabbahey]

Well plug them into tan(3x)=-sqrt(3) and use a calculator if you need to. And yes, you should check since you are the one doing the problem.

 

[meeklobraca]

I don't mean that I don't have to check and that someone else should check for me, I mean that I am trying to learn how to do these problems, so I don't have to. I can look at a problem like that, figure it out, and be confident enough with the answer that I don't have to check.

 

[NoMoreExams]

I don't mean that I don't have to check and that someone else should check for me, I mean that I am trying to learn how to do these problems, so I don't have to. I can look at a problem like that, figure it out, and be confident enough with the answer that I don't have to check.

1) you're = you are, your means something different :)

2) you should ALWAYS check your answers regardless of how confiden you are in them