How Do You Solve Trigonometric Inequalities Involving Sin, Cos, and Tan?

[Physicsissuef]

Homework Statement

Find the values for x.

3sin^2x - 3sinxcosx + 2cos^2x > 1

Homework Equations

The Attempt at a Solution

3sin^2x - 3sinxcosx + 2cos^2x > 1

3sin^2x - 3sinxcosx + 2cos^2x > sin^2x+cos^2x

2sin^2x - 3sinxcosx + cos^2x > 0

What to do next?

 

[Kurret]

You can use the same formula sin^2+cos^2=1 again. Now rewrite the equation in terms of only one trigonometric function.

 

[quantumdude]

I think it would be easier to just factor it the way it is.

 

[Physicsissuef]

What to do now?
If I try with sin^2+cos^2=1[/tex]<br /> <br /> I would get:<br /> sin^2x-3sinxcosx&amp;gt;-1<br /> <br /> Tom Mattson:<br /> (sinx-cosx)(2sinx-cosx)&amp;gt;0<br /> I am wondering why in my book says, I am giving quote:<br /> <br /> <blockquote data-attributes="" data-quote="my textbook" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> my textbook said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> We have 2 cases, the 1-st one if cosx = 0 and the 2-nd if cosx \neq 0 </div> </div> </blockquote><br /> Hm...

 

[quantumdude]

Your book probably took a different approach. Doing it my way, the two cases are:

1.) \sin(x)-\cos(x)&gt;0 and 2\sin(x)-\cos(x)&gt;0, or

2.) \sin(x)-\cos(x)&lt;0 and 2\sin(x)-\cos(x)&lt;0.

 

[Physicsissuef]

But they have 3 solutions for x.
The first: x=\frac{\pi}{2}+k\pi
The second: tgx&lt;\frac{1}{2}
The third: tgx&gt;1

In your case we have only 2:
tgx&lt;\frac{1}{2}
tgx&gt;1

 

[quantumdude]

No, Physicsissuef. When you rearranged the inequalities, you divided by \cos(x). That means you obtained results that are valid only for \cos(x)\neq 0. To find out what happens when \cos(x)=0, you have to examine those points separately.

So, for which x is \cos(x)=0, and do the inequalities hold for all such x?

 

[Physicsissuef]

But both points for cosx=0 and \cos(x)\neq 0 are valid.
For pi/2 and -pi/2 cosx=0

 

[quantumdude]

But both points for cosx=0 and \cos(x)\neq 0 are valid.

That's exactly what I just told you. When you divided by \cos(x), you obtained results that are valid only for \cos(x)\neq 0. You then have to go back and find out if the inequalities hold for all x such that \cos(x)=0.

 

[Physicsissuef]

but why not sin(x)=0, that's my question, or why not sin(x)=1 or cos(x)=1, etc.etc, how will I know which of all of this values to choose?

 

[quantumdude]

but why not sin(x)=0,

You didn't divide by \sin(x), did you?

that's my question, or why not sin(x)=1 or cos(x)=1, etc.etc, how will I know which of all of this values to choose?

Question: What did you divide by?

Answer: \cos(x).

Question: So what are the only values of x for which that division is not defined?

Answer: The values of x for which \cos(x), the thing you divided by, is zero.

 

[Physicsissuef]

what means "the division is not defined"?
Do you mean \frac{sinx}{cosx}=tgx, so for cosx=0, there will be infinity, but if it is only cosx=0, there will be no infinity right?

 

[tiny-tim]

But they have 3 solutions for x.
The first: x=\frac{\pi}{2}+k\pi
The second: tgx&lt;\frac{1}{2}
The third: tgx&gt;1

Hi Physicsissuef!

Solutions 2 and 3 are in terms of tanx, and technically tanx is undefined at multiples of π/2.

For example, just below π/2, it's huge and positive, but just above π/2 it's huge and negative. So at π/2 it's both-infinity-and-minus-infinity … which doesn't help with tanx > 1, does it?

So they had to add Solution 1, just to cover all possibliities!