Solving for 2Cos(θ)+8Cos(θ)+4=0

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Homework Help Overview

The discussion revolves around solving the equation 2 cos(θ) + 8 cos(θ) + 4 = 0, with a focus on the interval 0 ≤ θ ≤ 2π. Participants are attempting to clarify the correct formulation of the problem and explore the implications of their findings.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants initially present a factorization approach but later identify potential typos in the equation, suggesting it should be 3 cos²(θ) + 8 cos(θ) + 4 = 0. There are discussions about the notation used in expressing solutions and the importance of clarity in mathematical writing.

Discussion Status

The conversation is ongoing, with participants providing feedback on each other's notation and reasoning. Some participants express confusion about the expectations for solving the problem, indicating a mix of graphical and analytical approaches. There is no explicit consensus on the correct method or final answers yet.

Contextual Notes

Participants mention that they have been instructed to find solutions graphically, which may influence their approach to the problem. There is an emphasis on the importance of clear mathematical communication, especially in the context of homework submissions.

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1.Solve 2 cos(theta)+8cos(theta)+4=0 for 0≤theta≤2π



Homework Equations





3. (3cosTheta+2)(cosTheta+2)=0
cosTheta=-2/3 = 2.3 cosTheta=-2(no solution)
Solutions are at 3.9 and 2.4


Would someone be kind enough to check please?
 
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lemon said:
1.Solve 2 cos(theta)+8cos(theta)+4=0 for 0≤theta≤2π



2. Homework Equations



3. (3cosTheta+2)(cosTheta+2)=0
cosTheta=-2/3 = 2.3 cosTheta=-2(no solution)
Solutions are at 3.9 and 2.4

Would someone be kind enough to check please?
Are you missing an exponent in your first equation?
Assuming that's the case, some of the work you show is confusing and incorrect.
lemon said:
cosTheta=-2/3 = 2.3
-2/3 [itex]\neq[/tex] 2.3 and cos(theta) [itex]\neq[/tex] 2.3<br /> What you should say is <br /> cos(theta) = -2/3 ==> theta = 2.3005 (rad).<br /> The other value is approx. 3.9827 (rad). <br /> When rounded to one decimal place, these are 2.3 and 4.0, which are different from the values you show.[/itex][/itex]
[itex][itex][/itex][/itex]
 
Sorry. There is a typo. The original problem is:
3cos - not 2 cos from the beginning of the problem.
 
There is probably another typo, too, since I think you meant 3cos2(theta) in your equation.
 
I'M not sure what you mean.
The original problem is as follows: in full:
3cos2Theta+8cosTheta+4-0
 
The original problem is as follows: in full:
3cos2Theta+8cosTheta+4-0
 
The original problem is as follows: in full:
3cos2Theta+8cosTheta+4=0
 
lemon said:
I'M not sure what you mean.
The original problem is as follows: in full:
3cos2Theta+8cosTheta+4-0

This is what you wrote:
lemon said:
Solve 2 cos(theta)+8cos(theta)+4=0 for 0≤theta≤2π
 
Yes it was a typo. The 2 should have been a 3
 
  • #10
Yes. Your right Mark. I totally messed that up. Excuse me.
It should have been the following. Let me start again.

3cos2x+8cosx+4 0≤x≤2π
(3cosx+2)(cosx+2)
cosx=-2/3 = 2.3 cox=-2 (no solution)

solutions are at: 2.3 and 2π-2.3=3.98

How's that?
 
  • #11
Yes, putting x= 2.3 into the equation gives .00156 which is correct to two significant figures which is what you appear to be using.
 
  • #12
lemon said:
3cos2x+8cosx+4 0≤x≤2π
(3cosx+2)(cosx+2)
cosx=-2/3 = 2.3 cox=-2 (no solution)

solutions are at: 2.3 and 2π-2.3=3.98

How's that?
Just to reiterate what Mark said earlier...don't write

cos x = -2/3 = 2.3

-2/3 obviously doesn't equal 2.3, and cos x isn't equal to 2.3 either. It's very, very sloppy notation, and it probably annoys whoever's grading your homework.
 
  • #13
So, the only thing you wanted to comment on was, well, just to complain really?
 
  • #14
I'm sorry Hallsofivy. I don't understand what you mean by putting 2.3 into the equation. The solutions are already found on the cos graph at 2.3 and 3.98, aren't they?
Perhaps, my study and homework isn't going to the next stage, or something. Not sure. But this is all we have been asked/shown to do. Just to find the other solutions on the graphs, having the first.
 
  • #15
You're supposed to be solving the equation analytically, not by looking at a graph. What HallsOfIvy was saying was that x = 2.3 is close to the actual solution, since substituting that value into your original equation doesn't give 0, but a number that's reasonably close to 0. That solution is closer to 2.3005.
 
  • #16
ok. I understand. Thanks. But we haven't been shown that. We are told to just draw the graph and find the corresponding points/solutions. So, 2.3 would be the right answer for me. thanks for the help.
 
  • #17
lemon said:
So, the only thing you wanted to comment on was, well, just to complain really?
I'd say it's not so much a complaint than constructive criticism. What you wrote is just blatantly incorrect. I suspect you're thinking, "Who cares? You knew what I meant," but it's like writing a sentence full of grammatical errors. The reader might be able to figure out what you meant, but it still makes you look illiterate. On a more practical note, some graders will nail you for writing stuff like that. Many, especially in math courses, do care how you use the notation even though they know what you meant. Others just aren't willing to put the effort into figuring out what you meant; they'll just glance at what you wrote, decide it looks like nonsense, and mark you off.

It's not exactly difficult to write "cos x=-2/3 -> x=2.3"; it's just an arrow and an X more than what you wrote. So why not just write it correctly and clearly?
 
  • #18
Why not try to be a little more sensitive in your original "constructive" criticism. It's not exactly difficult to write a few damping words. It's just a "why don't you" or "perhaps it would be more appropriate" more than you wrote.
I will of course take it as constructive criticism, and duly thank you for your guidance.
Good night.
 
  • #19
lemon said:
We are told to just draw the graph and find the corresponding points/solutions. So, 2.3 would be the right answer for me.

It would have been useful for us to know that you were supposed to find the solutions graphically.
 
  • #20
lemon said:
Why not try to be a little more sensitive in your original "constructive" criticism. It's not exactly difficult to write a few damping words. It's just a "why don't you" or "perhaps it would be more appropriate" more than you wrote.
I will of course take it as constructive criticism, and duly thank you for your guidance.
Good night.

Vela's criticism should be much more welcoming to you than a grader's response would be. If I were to write what you did I wouldn't expect anything more than for it to be wrong.

Look at it again:

cosx=-2/3=2.3

This suggests cosx=-2/3, cosx=2.3 and -2/3=2.3
Obviously it's not right and skipping on writing an extra few symbols, specifically x=2.3, suggests laziness on your behalf. Be sure to develop proper habits early on :-p
 
  • #21
Vela. I understand and can see your point. I apologize for my bad habits and will do better to consider my writings more in the future.
But please consider that just entering into a thread without being known and making criticisms can seem very harsh. Receiving words by text communication doesn't carry with it the subtle nuances that face to face communication does, making it cold and hard in its nature; and I'm generally not an over sensitive person. Had we met before here, then it would not have mattered to me. Try to imagine your feeling if somebody you had never met before suddenly came into your office and said something like "Hay you! Stop doing what your doing in that sloppy manner". You'd probably feel a little insulted and invaded.
I know what you do here is great and you are very appreciated by students such as I. Your efforts make an immense difference to our learnings and the last thing that I wish is to insult anybody trying to help me.
I will do better and hope our relations can improve.
 

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