Solving for x in the Inequality cos(x) ≤ 5/3

In summary, the conversation discusses a confusion about an inequality involving cos(x) and a number. The question specifically asks for the solution in interval notation within a given range. The confusion arises from the fact that the number in the inequality is greater than 1, which is not possible for cos(x) as its range is [-1,1]. However, it is explained that the inequality is still valid in this case and has a solution for the entire given range. It is suggested that this may be a typo or oversight, but ultimately the solution is to accept the inequality as is and provide the answer within
  • #1
Vital
108
4

Homework Statement


Hello!

The task is to express the exact answer in interval notation, restricting your attention to -2π ≤ x ≤ 2π.

Homework Equations


The given inequality:
cos(x) ≤ 5/3

The Attempt at a Solution



I have only one doubt here, and I don't see my mistake.
I see that if cos(x) was strictly less than (namely <) 5/3, then the answer is [-2π, 2π].

But the inequality says that cos(x) is less or equal to 5/3, but cos(x) can't be equal to any number larger than 1, because its range is [-1, 1]. Why then this inequality is considered having any solutions at all (because of this "equal to" sign)?

Thank you!
 
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  • #2
≤ means "less than or equal to." If a < b is true, then a ≤ b is necessarily true.
 
  • #3
DrClaude said:
≤ means "less than or equal to." If a < b is true, then a ≤ b is necessarily true.
Yes, of course. That is why I am curious to see my mistake. My question is not about how to interpret the ≤ sign, but why it is there at all: as I said in my post, cos(x) cannot be bigger than 1, and 5/3 is obviously bigger than 1.
 
  • #4
I don't see why you are bothered by the ≤ but not by the 5/3. They are both arbitrary choices.
 
  • #5
DrClaude said:
I don't see why you are bothered by the ≤ but not by the 5/3. They are both arbitrary choices.
Well, because if the inequality states that cos(x) has to be less or equal to some number, then both choices should be valid; but in this case the choice of equality is invalid, hence the whole expression should be invalid. Why this presumption is incorrect?
 
  • #6
Vital said:
Well, because if the inequality states that cos(x) has to be less or equal to some number, then both choices should be valid
No, this is exactly what I pointed out above: it is or, not and. (And it has to be or, otherwise it could never be true.) 1 ≤ 2 is a true statement.
 
  • #7
DrClaude said:
No, this is exactly what I pointed out above: it is or, not and. (And it has to be or, otherwise it could never be true.) 1 ≤ 2 is a true statement.
All right. :-) Thank you very much!
 
  • #8
Sure you copied the question out right? cos (x) ≤ 3/5 is a more likely sounding condition for an exercise on.
 
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  • #9
epenguin said:
Sure you copied the question out right? cos (x) ≤ 3/5 is a more likely sounding condition for an exercise on.
Glad you said that. No, of course, I copied it correctly, and hence my confusion. I also thought that cos(x) ≤ 3/5 sounds more logical here, and, honestly, I still think that cos(x) ≤ 5/3 is invalid and has no solutions, even if it we should read ≤ as a logical OR.
 
  • #10
Vital said:
I still think that cos(x) ≤ 5/3 is invalid and has no solutions, even if it we should read ≤ as a logical OR.
Better get that thought out of your mind. At one point, you have to accept things as they here, even if it is just a question of convention or speaking the same language as everyone else.
 
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  • #11
DrClaude said:
Better get that thought out of your mind. At one point, you have to accept things as they here, even if it is just a question of convention or speaking the same language as everyone else.
:-)
 
  • #12
Vital said:
Glad you said that. No, of course, I copied it correctly, and hence my confusion. I also thought that cos(x) ≤ 3/5 sounds more logical here, and, honestly, I still think that cos(x) ≤ 5/3 is invalid and has no solutions, even if it we should read ≤ as a logical OR.

OK, often people here bang their heads against questions which later turn out to be students' mistranscriptions so I checked. But still in about a quarter of the remaining cases when the Profs setting the questions have been asked they say it was, or must have been, a typo or oversight. :oldbiggrin:

I'd say it's not that your inequality has no solution, rather that it is true for the whole of the given range.

If this is homework depending on situation and personality you might supplement yr answer adding 'if the question were ≤ 3/5 then ... ≤ x ≤... '...

It was reasonable, even creditable, to be puzzled.
 
  • #13
epenguin said:
OK, often people here bang their heads against questions which later turn out to be students' mistranscriptions so I checked. But still in about a quarter of the remaining cases when the Profs setting the questions have been asked they say it was, or must have been, a typo or oversight. :oldbiggrin:

I'd say it's not that your inequality has no solution, rather that it is true for the whole of the given range.

If this is homework depending on situation and personality you might supplement yr answer adding 'if the question were ≤ 3/5 then ... ≤ x ≤... '...

It was reasonable, even creditable, to be puzzled.

Thank you. Well, no, I don't have any "authority" to "report" to as I study on my own, and this was one of exercises from the textbook.
 
  • #14
What date and what edition was the textbook? Authors are glad to receive queries about possible misprints or things that may be unintended or misleading.
 
  • #15
epenguin said:
What date and what edition was the textbook? Authors are glad to receive queries about possible misprints or things that may be unintended or misleading.
2013
 
  • #16
Vital said:
Glad you said that. No, of course, I copied it correctly, and hence my confusion. I also thought that cos(x) ≤ 3/5 sounds more logical here, and, honestly, I still think that cos(x) ≤ 5/3 is invalid and has no solutions, even if it we should read ≤ as a logical OR.

Why would you think that ##\cos(x) \leq 5/3## has no solutions?

The standard definition of ##a \leq b## is "##a## is less than ##b##, or ##a## is equal to ##b##". That is the universally-accepted definition of "##\leq##", and nobody (you included) has the right to change it. It does NOT say "##a## is less than ##b## and ##a## equals ##b##", because that would be impossible, no matter what values you choose for ##a,b##. It is absolutely true that ##1 \leq 5/3##, or ##1 \leq 10,000,000##; in fact ## \leq c## is true for any constant ##c## that is 1 or larger.
 

Related to Solving for x in the Inequality cos(x) ≤ 5/3

What is the definition of an inequality?

An inequality is a mathematical statement that compares two quantities using symbols such as <, >, ≤, or ≥. It indicates that one quantity is less than, greater than, less than or equal to, or greater than or equal to the other quantity.

What does "solving for x" mean?

Solving for x means finding the value or values of x that make the inequality true. In other words, we are looking for the set of solutions that satisfy the given inequality.

How do I solve for x in an inequality with cosine?

To solve for x in an inequality with cosine, we need to use inverse cosine or arccosine on both sides of the inequality. This will help us isolate x and determine its value. Keep in mind that the solution may involve multiple steps and may require the use of trigonometric identities.

What does it mean when cosine is less than or equal to a fraction?

When cosine is less than or equal to a fraction, it means that the cosine function is equal to or less than that fraction for some or all values of x. This can also be interpreted as the cosine values being within a certain range of values, depending on the given fraction.

Are there any restrictions when solving for x in this inequality?

Yes, there are restrictions when solving for x in this inequality. Since cosine is a periodic function, there will be infinitely many solutions. However, we can restrict the domain of the inequality to a specific interval or range of values to find a finite set of solutions.

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