Question

question

Answer 1

Can you help me with a question?

Answer 2

Find the minimum value of \(\dfrac{2}{\sin x}+3\sin x\)

Answer 3

IDK what that is. What grade?

Answer 4

Have you considered the 1st Derivative?

Answer 5

i m not allowed to use calculos

Answer 6

@Catch.me

Answer 7

allowed to use AM-GM?

Answer 8

yes

Answer 9

See, this is an example of why we need to see YOUR work in the VERY FIRST post. Where are you? What are you doing? What are the constraints?

Answer 10

i only ask for help when i m stuck

Answer 11

You entirely miss the point. SHOW YOUR WORK. No excuses. Try Something. How doe AM-GM apply to this problem? You didn't eve tell us calculus was not allowed. You could have told us that.

Answer 12

Well the original was to find minimum of this value \(4\csc^{2} x+9\sin^{2} x\)

which i was stucked at this part \(\left(\dfrac{2}{\sin x}-3\sin x\right)^{2}+12\)

I have also helped a considerable amount of people in this site,
and what i expect is path to be shown leading to the answer and not the answer
but considering u are a moderator , i didn't expect u to be so rude.

Answer 13

here is my reasoning:
the minimum value of such function occurs at point x, such that the subsequent condition holds:
\[\Large \frac{2}{{\sin x}} = 3\sin x\]
Please keep in mind that I'm searching for the minimum value inside this interval: \((0, \pi)\)
Now, from that equation we get:

\[\sin x = \sqrt {\frac{2}{3}} \]
which is the positive solution only

Answer 14

and the requested minimum value is:
\[\Large \frac{2}{{\sin x}} + 3\sin x = 2 \cdot \sqrt {\frac{3}{2}} + 3 \cdot \sqrt {\frac{2}{3}} \simeq 4.8989\]

Answer 15

I'm never rude. Honesty and directness are often mistaken for rudeness.

Just show your work. You had plenty to show. Don't make anyone drag it out of you.

Answer 16

I am not pointing at anyone's ability , he has helped me before and i appreciate it , but as a moderators says something to a user asking for help in a a fashion that seems a little threatening then the existing users and new users can inhabit a lack of faith in the site , a student expects the teacher to be polite if teacher doesn't wanna help that's ok. The sense of disrespect or rudeness can make the users have a bad reputation of the site .

Answer 17

I like how tkhunny is blunt, to be honest. Let's face facts, in college, in higher education, in the workplace, we learn the most out of the direct people not the people that honey their words. A teacher too focused on semantics and being polite is not a teacher, that is a politician.

In particular, tkhunny helps represent the volunteering nature of this site. Frankly, I don't help nearly as often as I used to. Why? Users these days are entitled and act like they deserve the help for free. They take it for granted. Maybe they deserved to be helped, but with that kind of attitude, I think the help they need is learning how to figure things on their own. Teachers have ALWAYS had more discretion than students on teaching methods. Why? They have been there longer (seniority) and they are generally more qualified (better discretion). While tkhunny might seem like an extreme, frankly, a user should always be grateful to recieve help, whether or not it is blunt or not, whether or not it even seems helpful or not.

Answer 18

I also want to make a point that might not be true ,
After tkhunny made that post some users who were trying to help me before went away from this question one fear might be they just might have thought that if they helped me then the tkhunny might have taken an action against them.

Answer 19

No action ever has been threatened, nor should be feared, for trying to help.

@Michele_Laino I wasn't seeing this at all. It finally dawned on me that everything was positive. Good work.

Answer 20

no problem :) and thanks!! :) @tkhunny

Answer 21

@mathmath333 hope not so late
I will do it slowly
to get min for \[\frac{ 2 }{ \sin x } + 3 \sin x \]
we will have \[\frac{ 2 }{ \sin x } + 3 \sin x -D = 0 \] where D translation on y axis as it would only cross x at only one point
\[\sin x \neq 0\] because you know first term will diverge.
now we have \[3 \sin ^{2} x - D \sin x + 2 = 0\]
it's a min so it will have only one root or one x cross giving \[3 (\sin x - q)^{2} = 0\]
just expand and equate coefficient for both equations
\[-6q = - D\]
\[3q ^{2} = 2\] hence \[q = \sqrt{\frac{ 2 }{ 3 }}\]
and min. value \[D = 6\sqrt{\frac{ 2 }{ 3 }} =4.8989....... \]
agreed with @Michele_Laino ,but I didn't understand his reasoning.
source: http://mathforum.org/library/drmath/view/62730.html
I remember that I saw this method before with math history (maybe not) anyway Dr.Peterson did a great job.
I think it would be tricky to do this with AM-GM, but I will try(actually I am very sleep maybe tomorrow)

Answer 22

@tkhunny Was that legal!!

Answer 23

\(\bf \Huge \color{black}{\ddot\smile}\)