this should work, no?

A

allegansveritatem

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I spent two hours on this equation today trying to get it into some form I could use to get the zeros. After each transformation I would test the result in the calculator to see if I got the right intercepts. I seem to have learned, among other things that the term sinxcosx in not divisible into sinx and cosx. Here is one attempt to do that. It absolutely failed to serve my turn:
why1.PNG

Why doesn't this work? I followed all the rules of algebra...so maybe the rules are different when trig functions are in the mix?
 
In what sense doesn't it work? Do you mean that you just couldn't get to a solution from here? Or that you found graphically that it is not equivalent to the original? Or something else?

I would avoid dividing by anything with a variable, which I think you have been told already (unless that was someone else). I'd take it from the first line (where something equals zero) and factor. Then I'd observe that one of the factors can be rewritten to involve only one function, namely the cosine, so I'd do that. These are two very useful tools.
 
[math] \dfrac {\sin(x)}{\sin(x)} \neq 1[/math]. You were already told this! Rather [math] \dfrac {\sin(x)}{\sin(x)}[/math] is a piecewise function. Do you know which one?
 
Dr.Peterson said:
In what sense doesn't it work? Do you mean that you just couldn't get to a solution from here? Or that you found graphically that it is not equivalent to the original? Or something else?

I would avoid dividing by anything with a variable, which I think you have been told already (unless that was someone else). I'd take it from the first line (where something equals zero) and factor. Then I'd observe that one of the factors can be rewritten to involve only one function, namely the cosine, so I'd do that. These are two very useful tools.
Click to expand...
I want to say that I was sitting quietly a little while ago and it suddenly hit me that I had posted something like this just a few days ago and learned that it was unwise because the variable could well be a zero. This was just 2 or 3 days ago! How dumb is that? I sometimes get on a plane that flies on autopilot straight to the land of the Flat Earthers. Anyway, I will come back later to go over the posts but I just had to come now and set this down.
 
Jomo said:
[math] \dfrac {\sin(x)}{\sin(x)} \neq 1[/math]. You were already told this! Rather [math] \dfrac {\sin(x)}{\sin(x)}[/math] is a piecewise function. Do you know which one?
Click to expand...
I will come back to your post later tonight. I can say right now that I without thinking about it I have no idea. I do recall something about piece wise functions but even for that I will need some brushing up. I will return.
 
allegansveritatem said:
I will come back to your post later tonight. I can say right now that I without thinking about it I have no idea. I do recall something about piece wise functions but even for that I will need some brushing up. I will return.
Click to expand...
x/x is NOT always 1. When is it 1? When is it not 1? When it is not 1 then what is it?
 
Jomo said:
[math] \dfrac {\sin(x)}{\sin(x)} \neq 1[/math]. You were already told this! Rather [math] \dfrac {\sin(x)}{\sin(x)}[/math] is a piecewise function. Do you know which one?
Click to expand...
I would say that there are two categories: x=0 and x does not =zero. When x = zero then the function is undefined. When x equals anything other than zero then sinx/sinx does = 1. No?
 
Jomo said:
x/x is NOT always 1. When is it 1? When is it not 1? When it is not 1 then what is it?
Click to expand...
AS I just wrote in reply to your earlier post, when x is = to anything but zero then x/x is = to 1. When x is zero, then the game is up and the function is undefined. By the way, can you suggest some approach to solving the equation in my original post? I am having a hard time getting that pesky cosx to fit in a factor. Maybe I should shoot for cosx based factors...lbut I seem to have tired everything without luck. I know already what the solutions are, namely minus 2pi, minus pi, 0, pi and 2pi. I know this becasue I truned the original equation into a function and mechanically graphed it. So therelfore sinx = 0...but how to demonstrate the process whereby we know this?
 
Sinx cosx-2 sin^3x cos x - sinx =0
Sin x( cos x -2sin^2x cosx-1)=0
Then replace the 2 sin^2x with an expression using cos
 
allegansveritatem said:
I spent two hours on this equation today trying to get it into some form I could use to get the zeros. After each transformation I would test the result in the calculator to see if I got the right intercepts. I seem to have learned, among other things that the term sinxcosx in not divisible into sinx and cosx. Here is one attempt to do that. It absolutely failed to serve my turn:
View attachment 18442

Why doesn't this work? I followed all the rules of algebra...so maybe the rules are different when trig functions are in the mix?
Click to expand...
sin(x) * cos(x) * [1 - 2sin^2(x)] - sin(x) = 0

sin(x) * { cos(x) * [ 2 * cos^2(x) - 1] - 1} = 0

so, either

sin(x) = 0 \(\displaystyle \ \ \to \) x = n * \(\displaystyle \pi \)

or

cos(x) * [ 2 * cos^2(x) - 1] - 1 = 0

cos(x) * cos(2x) = 1 \(\displaystyle \ \ \to \) x = 2 * n * \(\displaystyle \pi \)
 
Subhotosh Khan said:
sin(x) * cos(x) * [1 - 2sin^2(x)] - sin(x) = 0

sin(x) * { cos(x) * [ 2 * cos^2(x) - 1] - 1} = 0

so, either

sin(x) = 0 \(\displaystyle \ \ \to \) x = n * \(\displaystyle \pi \)

or

cos(x) * [ 2 * cos^2(x) - 1] - 1 = 0

cos(x) * cos(2x) = 1 \(\displaystyle \ \ \to \) x = 2 * n * \(\displaystyle \pi \)
Click to expand...
Thanks. I will try this. Both suggestions look like variants of paths I considered but got waylaid in yesterday. But I see here where I might have gone astray.
 
allegansveritatem said:
AS I just wrote in reply to your earlier post, when x is = to anything but zero then x/x is = to 1. When x is zero, then the game is up and the function is undefined. By the way, can you suggest some approach to solving the equation in my original post? I am having a hard time getting that pesky cosx to fit in a factor. Maybe I should shoot for cosx based factors...lbut I seem to have tired everything without luck. I know already what the solutions are, namely minus 2pi, minus pi, 0, pi and 2pi. I know this becasue I truned the original equation into a function and mechanically graphed it. So therelfore sinx = 0...but how to demonstrate the process whereby we know this?
Click to expand...
Is sin(x) = 0? If yes, then sin(x) is one solution and now you can divide by six.
If no, then sin(x) is not a solution and you can divide by sin(x).

As Dr Peterson suggested you should try not to divide by unknown values. But if you insist then you must check like I suggested above.
 
Jomo said:
Is sin(x) = 0? If yes, then sin(x) is one solution and now you can divide by six.
If no, then sin(x) is not a solution and you can divide by sin(x).

As Dr Peterson suggested you should try not to divide by unknown values. But if you insist then you must check like I suggested above.
Click to expand...
What?!
 
Jomo said:
Is sin(x) = 0? If yes, then sin(x) is one solution and now you can divide by six.
If no, then sin(x) is not a solution and you can divide by sin(x).

As Dr Peterson suggested you should try not to divide by unknown values. But if you insist then you must check like I suggested above.
Click to expand...
Divide by six? How do you come by that?
 
so, I went round to work again to day and came up with the following which, being constrained by notebook runover (?), I will present in two images:
(I apologize for not numbering the steps in image two. Forgot.
sincos05-041.PNG
sincos05-04.PNG
I am a little mistrustful of my results in the second page. Seems like that factor is a pretty complex hombre as factors go ( in my experience anyway) but it seems to work out and things fell out so that the cosx figure was impossible and had to be crossed off. Is there anything wrong with my algebra in the second page? I know the first page is all good.
 
On your 2nd page from line 2 to line 3 is a mistake. You had a product which you lost. Try again.

From line 5 to 6 why would subtraction become multiplication. Is 9-3 = 9*(-3)? NO!

6 really should know what 6/2 equals!
 
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