this should work, no?

Jomo said:
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!
Click to expand...
I will redo the thing and post results tomorrow. Yes, the 6/2 thing was pretty coy, I admit. Thanks for checking it out.
 
Jomo said:
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!
Click to expand...
so, here is my latest attempt at getting a value for cosx:
cos0505.PNG
I know I am correct up to the question mark, and then I feel like a dog that has chased down a car and when it has cornered the car, it doesn't know what to do with it.
 
On line 3, you misplaced a parenthesis, but the next line is correct.

The factoring on line 7 is useless; in solving an equation, we factor when the other side is zero, because then at least one of the factors is known to be 0. When the other side is 1, it tells you nothing. Go back to line 6 to continue.

I would now make a substitution: u = cos(x).
 
Dr.Peterson said:
On line 3, you misplaced a parenthesis, but the next line is correct.

The factoring on line 7 is useless; in solving an equation, we factor when the other side is zero, because then at least one of the factors is known to be 0. When the other side is 1, it tells you nothing. Go back to line 6 to continue.

I would now make a substitution: u = cos(x).
Click to expand...
I know it didn't make much sense to factor at that point, but I was game to try anything. I will try what you suggest. I already know what I should get, but I have yet to demonstrate the fact algebraically.
 
Dr.Peterson said:
On line 3, you misplaced a parenthesis, but the next line is correct.

The factoring on line 7 is useless; in solving an equation, we factor when the other side is zero, because then at least one of the factors is known to be 0. When the other side is 1, it tells you nothing. Go back to line 6 to continue.

I would now make a substitution: u = cos(x).
Click to expand...
so, I attacked it again today and finally came up with this:
0506.PNG
it seems that algebra and reason got together and with trigonometry as midwife brought this forth. Good? Bad? I don't know but it was all I could come up with.
 
Good work, mostly ...

At this point, the problem gets ugly.

How did you conclude that u has to be 1?

By inspection, we can say that u can be 1; but you need to confirm that it is the only solution. You might just graph it and see; or you might factor (u-1) from the cubic polynomial and see that the quadratic that remains has no real solutions.

That's an unpleasant way for a problem to end, so I have long wondered if the problem has been copied wrong at some stage; but it is (part of) the right answer. (I say "part of" because we left the other factor, sin x, lying in the dust long ago, and must not forget the solution that provides.)

But now you have to tell us whether there was any restriction on the domain of x, and whether x=0 is the only solution to cos x =1!
 
Dr.Peterson said:
Good work, mostly ...

At this point, the problem gets ugly.

How did you conclude that u has to be 1?

By inspection, we can say that u can be 1; but you need to confirm that it is the only solution. You might just graph it and see; or you might factor (u-1) from the cubic polynomial and see that the quadratic that remains has no real solutions.

That's an unpleasant way for a problem to end, so I have long wondered if the problem has been copied wrong at some stage; but it is (part of) the right answer. (I say "part of" because we left the other factor, sin x, lying in the dust long ago, and must not forget the solution that provides.)

But now you have to tell us whether there was any restriction on the domain of x, and whether x=0 is the only solution to cos x =1!
Click to expand...
yes, there is a restriction: =2pi to 2pi interval. Sinx = 1 =0 and when I look at the graph I see that x axis is crossed 5 times, -2pi, -pi, 0 , pi, 2pi. As for u = 1 being the only solution...if the left side equals one, how can it be anything else?
 
Cubic equations typically have three solutions. The fact that the LHS is zero for one value of x does not imply that is the only solution.

Here is the graph of [MATH]2u^3-u-1[/MATH]:

1588821684488.png

You can see that it is zero only for u=1; but it wouldn't take much for it to have two or three solutions.
 
Dr.Peterson said:
Cubic equations typically have three solutions. The fact that the LHS is zero for one value of x does not imply that is the only solution.

Here is the graph of [MATH]2u^3-u-1[/MATH]:

View attachment 18571

You can see that it is zero only for u=1; but it wouldn't take much for it to have two or three solutions.
Click to expand...
but...in the present situation where the expression is restricted, so to speak by = 1...I just don't see how any other solution could fit there. I will check into this and think about it and post back tonight. Thanks for the graph. I graphed the original equation but not this part of things.
 
allegansveritatem said:
but...in the present situation where the expression is restricted, so to speak by = 1...I just don't see how any other solution could fit there. I will check into this and think about it and post back tonight. Thanks for the graph. I graphed the original equation but not this part of things.
Click to expand...
In what sense the expression is restricted by =1?
How about 0.1u3-u-1? It has the same "restriction". But it has 3 roots:

y=.1u^3−u−1 - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.
www.wolframalpha.com www.wolframalpha.com
 
lev888 said:
In what sense the expression is restricted by =1?
How about 0.1u3-u-1? It has the same "restriction". But it has 3 roots:

y=.1u^3−u−1 - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.
www.wolframalpha.com www.wolframalpha.com
Click to expand...
when I say restricted to 1 I mean the equation: -u +u^3=1 There is only one number that u can represent in said situation and that is 1. No?
 
allegansveritatem said:
0. So, I guess u actually equals...the cube root of 1+u.
Click to expand...
No!

I start with

- u + u^3 = 1.........................................(1)

when I put u = 1, into the above equation, I get

-1 + 1^3 =1

0 =1 ........................ not possible

So u = 1 is NOT a solution of equation (1).
 
Subhotosh Khan said:
No!

I start with

- u + u^3 = 1.........................................(1)

when I put u = 1, into the above equation, I get

-1 + 1^3 =1

0 =1 ........................ not possible

So u = 1 is NOT a solution of equation (1).
Click to expand...
yes...but the equation in question is this: -u+2u^3=1. In the case you are citing, i.e., -u+u^3=1 is not possible. Right. And in the case I have in mind and which I presented some time earlier 1 is possible and if we state that -u+2u^3 equals 1 then we of necessity must conclude that that u on the left side is 1. No? So in the equation you cite, you are right and in the one I cite, I am right. It's like that Dylan lyric, "You're right from your side and I'm right from mine."
 
allegansveritatem said:
if we state that -u+2u^3 equals 1 then we of necessity must conclude that that u on the left side is 1.
Click to expand...
See post #28. 1 is a solution. You can see from the graph that it's the only solution. But the equation itself does not make it obvious.
 
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