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Trigonometry inequalities

1. The problem statement, all variables and given/known data
Find the values for x.

[tex]3sin^2x - 3sinxcosx + 2cos^2x > 1[/tex]

2. Relevant equations



3. The attempt at a solution

[tex]3sin^2x - 3sinxcosx + 2cos^2x > 1[/tex]

[tex]3sin^2x - 3sinxcosx + 2cos^2x > sin^2x+cos^2x[/tex]

[tex]2sin^2x - 3sinxcosx + cos^2x > 0[/tex]

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

Tom Mattson

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I think it would be easier to just factor it the way it is.
 
What to do now?
If I try with [itex]sin^2+cos^2=1[/tex]

I would get:
[tex]sin^2x-3sinxcosx>-1[/tex]

Tom Mattson:
[tex](sinx-cosx)(2sinx-cosx)>0[/tex]
I am wondering why in my book says, I am giving quote:

my text book said:
We have 2 cases, the 1-st one if [itex]cosx = 0[/itex] and the 2-nd if [itex]cosx \neq 0[/itex]
Hm...
 

Tom Mattson

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

1.) [itex]\sin(x)-\cos(x)>0[/itex] and [itex]2\sin(x)-\cos(x)>0[/itex], or

2.) [itex]\sin(x)-\cos(x)<0[/itex] and [itex]2\sin(x)-\cos(x)<0[/itex].
 
But they have 3 solutions for x.
The first: [tex]x=\frac{\pi}{2}+k\pi[/tex]
The second: [tex]tgx<\frac{1}{2}[/tex]
The third: [tex]tgx>1[/tex]

In your case we have only 2:
[tex]tgx<\frac{1}{2}[/tex]
[tex]tgx>1[/tex]
 

Tom Mattson

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

So, for which [itex]x[/itex] is [itex]\cos(x)=0[/itex], and do the inequalities hold for all such [itex]x[/itex]?
 
But both points for [itex]cosx=0[/itex] and [itex]\cos(x)\neq 0[/itex] are valid.
For pi/2 and -pi/2 cosx=0
 

Tom Mattson

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

Tom Mattson

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but why not sin(x)=0,
You didn't divide by [itex]\sin(x)[/itex], 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: [itex]\cos(x)[/itex].

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

Answer: The values of [itex]x[/itex] for which [itex]\cos(x)[/itex], the thing you divided by, is zero.
 
what means "the division is not defined"?
Do you mean [itex]\frac{sinx}{cosx}=tgx[/itex], so for cosx=0, there will be infinity, but if it is only cosx=0, there will be no infinity right?
 

tiny-tim

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But they have 3 solutions for x.
The first: [tex]x=\frac{\pi}{2}+k\pi[/tex]
The second: [tex]tgx<\frac{1}{2}[/tex]
The third: [tex]tgx>1[/tex]
Hi Physicsissuef! :smile:

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? :frown:

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

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