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

[Vital]

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!

 

[DrClaude]

≤ means "less than or equal to." If a < b is true, then a ≤ b is necessarily true.

 

[Vital]

≤ 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.

 

[DrClaude]

I don't see why you are bothered by the ≤ but not by the 5/3. They are both arbitrary choices.

 

[Vital]

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?

 

[DrClaude]

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.

 

[Vital]

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!

 

[epenguin]

Sure you copied the question out right? cos (x) ≤ 3/5 is a more likely sounding condition for an exercise on.

 

[Vital]

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.

 

[DrClaude]

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.

 

[Vital]

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.

:-)

 

[epenguin]

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.

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.

 

[Vital]

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.

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.

 

[epenguin]

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.

 

[Vital]

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

 

[Ray Vickson]

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.