# sine inequality

• June 28th 2008, 06:11 AM
fardeen_gen
sine inequality
The set of all x in the interval [0, π] for which 2 sin^2 x - 3 sin x + 1 ≥ 0 is?

I got three values of x:
π/6, 5π/6 and π/2

Options for the question are:
A) [π/6, 5π/6] U {π/2}
B) [π/6, 5π/6] - {π/2}
C) [π/6, 5π/6]
D) None

What is the correct option? And can someone explain the methodology behing representing the values in an interval or set?
• June 28th 2008, 06:52 AM
mr fantastic
Quote:

Originally Posted by fardeen_gen
The set of all x in the interval [0, π] for which 2 sin^2 x - 3 sin x + 1 ≥ 0 is?

I got three values of x:
π/6, 5π/6 and π/2

Options for the question are:
A) [π/6, 5π/6] U {π/2}
B) [π/6, 5π/6] - {π/2}
C) [π/6, 5π/6]
D) None

What is the correct option? And can someone explain the methodology behing representing the values in an interval or set?

Consider $2u^2 - 3u + 1 \geq 0$ (where $u = \sin x$) .....

So the solution is found from $\sin x \leq \frac{1}{2}$ or $\sin x \geq 1$ .....

So the solution is $0 \leq x \leq \frac{\pi}{6} \, \cup \, x = \frac{\pi}{2} \, \cup \, \frac{5 \pi}{6} \leq x \leq \pi$.

Option D.
• June 28th 2008, 09:17 AM
Mathstud28
Quote:

Originally Posted by fardeen_gen
The set of all x in the interval [0, π] for which 2 sin^2 x - 3 sin x + 1 ≥ 0 is?

I got three values of x:
π/6, 5π/6 and π/2

Options for the question are:
A) [π/6, 5π/6] U {π/2}
B) [π/6, 5π/6] - {π/2}
C) [π/6, 5π/6]
D) None

What is the correct option? And can someone explain the methodology behing representing the values in an interval or set?

If you are asking what specifically these intervals represent.

The notation

$[a,b]\cup[c,d]$

Means

$\forall{x}\in[a,n]\wedge[c,d]$

Or in other words

"Every x that is contained in the set $[a,b]$ AND $[c,d]$"

------------------
The notation

$[a,b]-d$

means

"Every x contained with in the set $a,b]$ EXCEPT $d$"
• June 28th 2008, 09:21 AM
Moo
Quote:

Originally Posted by Mathstud28
If you are asking what specifically these intervals represent.

The notation

$[a,b]\cup[c,d]$

Means

$\forall{x}\in[a,n] {\color{red}\wedge} [c,d]$ << ${\color{red}\vee}$, not ${\color{red}\wedge}.$

Or in other words

"Every x that is contained in the set $[a,b]$ AND $[c,d]$" << should be OR

Quote:

Originally Posted by Mathstud28
The notation

$[a,b]-{\color{red}\{}d{\color{red}\}}$

means

"Every x contained with in the set $[a,b]$ EXCEPT $d$"

tsssk
• June 28th 2008, 09:24 AM
Mathstud28
Quote:

Originally Posted by Moo
huh

You've never seen that? He has it above...

Like

$A\cup{C}-D$?

That is what I have always seen in Set Theory books.
• June 28th 2008, 09:26 AM
Mathstud28
Quote:

Originally Posted by Moo
tsssk

My bad...I knew it was or...

Because it means all the elements of two sets that are included in either. So if an element is included in one set and not the other it is still included in the union of the two sets.
• June 28th 2008, 09:26 AM
Moo
Quote:

Originally Posted by Mathstud28
You've never seen that? He has it above...

Like

$A\cup{C}-D$?

That is what I have always seen in Set Theory books.

For this notation, yes, but not when dealing with such intervals. d represents a single element, so in general (dunno what your book says.) you have to put it in brackets {d}.

Quote:

Originally Posted by Mathstud28
My bad...I knew it was or...

Because it means all the elements of two sets that are included in either. So if an element is included in one set and not the other it is still included in the union of the two sets.

Of course... you know everything, but you don't take enough care to read yourself............... when someone tells you the solution, it's always "I would have done this way !" "i knew it was that !"

Tsssk
• June 28th 2008, 09:30 AM
Mathstud28
Quote:

Originally Posted by Moo
For this notation, yes, but not when dealing with such intervals. d represents a single element, so in general (dunno what your book says.) you have to put it in brackets {d}.

Plus, I edited my post, to correct your blunders.

Yes, Moo...I like how when I have a typo its a "blunder" but when somone else does it "N'inquiete pas, tout le monde peut se tromper" (Smirk)