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 [itex]sin^2+cos^2=1[/tex]<br /> <br /> I would get:<br /> $$sin^2x-3sinxcosx>-1$$<br /> <br /> Tom Mattson:<br /> $$(sinx-cosx)(2sinx-cosx)>0$$<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...[/itex]

 

[quantumdude]

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

1.) $\sin(x)-\cos(x)>0$ and $2\sin(x)-\cos(x)>0$, or

2.) $\sin(x)-\cos(x)<0$ and $2\sin(x)-\cos(x)<0$.

 

[Physicsissuef]

But they have 3 solutions for x.
The first: $$x=\frac{\pi}{2}+k\pi$$
The second: $$tgx<\frac{1}{2}$$
The third: $$tgx>1$$

In your case we have only 2:
$$tgx<\frac{1}{2}$$
$$tgx>1$$

 

[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: $$tgx<\frac{1}{2}$$
The third: $$tgx>1$$

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!