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