Solving for 2Cos(θ)+8Cos(θ)+4=0

  • Thread starter lemon
  • Start date
But thanks again. I will try not to make any more "lazy" mistakes. It's just that I'm not in a class situation and I just help my daughter with her homework at home, so we are not aware of all the rules and the etiquette.
  • #1
lemon
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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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  • #2
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
What you should say is
cos(theta) = -2/3 ==> theta = 2.3005 (rad).
The other value is approx. 3.9827 (rad).
When rounded to one decimal place, these are 2.3 and 4.0, which are different from the values you show.
 
  • #3
Sorry. There is a typo. The original problem is:
3cos - not 2 cos from the beginning of the problem.
 
  • #4
There is probably another typo, too, since I think you meant 3cos2(theta) in your equation.
 
  • #5
I'M not sure what you mean.
The original problem is as follows: in full:
3cos2Theta+8cosTheta+4-0
 
  • #6
The original problem is as follows: in full:
3cos2Theta+8cosTheta+4-0
 
  • #7
The original problem is as follows: in full:
3cos2Theta+8cosTheta+4=0
 
  • #8
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
 
  • #9
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.
 

FAQ: Solving for 2Cos(θ)+8Cos(θ)+4=0

What is the equation "2Cos(θ)+8Cos(θ)+4=0" used for?

The equation "2Cos(θ)+8Cos(θ)+4=0" is used to solve for the value of θ in trigonometric functions. It is commonly used in mathematics, physics, and engineering to solve problems involving angles and oscillatory systems.

What is the process for solving "2Cos(θ)+8Cos(θ)+4=0"?

To solve "2Cos(θ)+8Cos(θ)+4=0", you can use the trigonometric identity Cos(θ)+Cos(θ)=2Cos(θ) to rewrite the equation as 2Cos(θ)+8Cos(θ)+4=0. Then, factor out the common term of 2Cos(θ) to get 2Cos(θ)(1+4)+4=0. Simplify to get 2Cos(θ)(5)+4=0. Finally, divide both sides by 2 to isolate Cos(θ) and solve for θ using a calculator or table of values.

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What are the solutions to "2Cos(θ)+8Cos(θ)+4=0"?

The solutions to "2Cos(θ)+8Cos(θ)+4=0" are the values of θ that satisfy the equation. In this case, there are two solutions: θ=arccos(-2/5) and θ=arccos(-4/5). These can be found using a calculator or table of values.

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If you cannot find the exact solutions to "2Cos(θ)+8Cos(θ)+4=0", you can use a calculator or computer program to approximate the solutions. You can also use a graphing calculator to graph the equation and find the x-intercepts, which represent the solutions to the equation.

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