Finding the Number of Solutions in [0,2pi) for log5(cos x) = -1/2

So I'm still lost on what you were thinking, but I'm glad you got the right answer! So … in summary, the equation -log5(cos(x)) = -1/2 has two solutions in the interval [0,2pi): 63.43 and 296.57 degrees.
  • #1
Neophyte
44
0

Homework Statement


-log5(cos(x)) = -1/2

How many answers are there in the interval [0,2pi)

The Attempt at a Solution


5^-1/2 = cos(x)
1/sq rt 5 = cos(x)
sq rt 5/5 = cos(x)
0.447 = cos(x)
x = 63.43

Basically, I have no clue how to find the number of answers in the interval ; (
I think I have missed this question or a variation of it like 12 times now ; /.. Things I have tried that work sometimes ( good guessing... Unfortunately when I tried applying it to a new problem it didnt work) is seeing how many times 63 went into 180, simply assuming its two as 180-63, then seeing how many times it went into 360 but this was not even an answer choice so idk...
If the answer is 4 I will have to cry.
 
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  • #2
Hi Neophyte! :smile:

(have a pi: π and a degree: º and a square-root: √ and try using the X2 tag just above the Reply box :wink:)

Yes, cos(180º - 63º) = minus cos63º.

Just draw the graph of cos from 0 to 2π (0º to 360º) …

then draw a horizontal line anywhere across it …

what appears to be the rule for the points where it crosses? :wink:
 
  • #3
So both sin and cos are either two or zero (und) ? Is this the same rule that applies to triangles ? Such as if angle C is 30º and then your suppose to find angle A which is 60º; I thought it was zero because there was something about an angle not being bigger then another angle (lol) needless to say there are a lot of new rules and Idk when to apply them. [A probability game ; (] Basically I think the answer was 60º and 120º so there was two but when would you apply that is has to be less than some angle.
 
  • #4
Hi Neophyte! :smile:
Neophyte said:
So both sin and cos are either two or zero (und) ?

Yes (or one, if cos or sin = ±1, of course).

(what's und mean? :confused:)

Sorry, but I don't understand the rest of your post. :redface:

Anyway … the answers are 63º and … ? :smile:​
 
  • #5
tiny-tim said:
Anyway … the answers are 63º and … ? :smile:​

:rolleyes:. Well, if it is on the graph it seems it would be 297º. Isnt it restricted from 0:π so it would be one?

2. If sine were 22.2º it would be 22.2º and 337.8º (two) ?

3. If that part is correct, what I was trying to say above; :confused: the only difference between that and a triangle is you would subtract it from 180º? Oh, I think I know what the problem was :shy: the angles just had to when added be under 180º :smile:.

Thank you for your help by the way, I greatly appreciate it.

Ignore this When you had said add a X2 tag I was a bit confused and tried adding it to the page tags but too short, so I added two tags but now I see what you were talking about.:blushing:

But (und) was suppose to be undefined probably inaccurate though.
 
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  • #6
Neophyte said:
:rolleyes:. Well, if it is on the graph it seems it would be 297º. Isnt it restricted from 0:π so it would be one?

Yes, but didn't the question say [0,2pi)?
2. If sine were 22.2º it would be 22.2º and 337.8º (two) ?

Nooo … sine is a different shape graph …

sin22º = sin … ? :wink:
 
  • #7
tiny-tim said:
Yes, but didn't the question say [0,2pi)?

Nooo … sine is a different shape graph …

sin22º = sin … ? :wink:

It did but would it not be irrelevant as arccos is restricted to 0:π?

It would be from -π/2:π/2 so it would be -22º (337.8)
Doh, So it would be 22º and 158º ? But that is not in the interval for arcsin :confused:

Do the arc limitations even play a role in this?
 
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  • #8
Neophyte said:
Do the arc limitations even play a role in this?

It isn't in the question …
-log5(cos(x)) = -1/2

How many answers are there in the interval [0,2pi)

Forget arccos! :smile:​
 
  • #9
Neophyte said:
It did but would it not be irrelevant as arccos is restricted to 0:π?

It would be from -π/2:π/2 so it would be -22º (337.8)
Doh, So it would be 22º and 158º ? But that is not in the interval for arcsin :confused:

Do the arc limitations even play a role in this?
"arc limitations" on the "arc" functions are set in order that they actually be functions! In other words, that there be only one value. In a question asking "how many values", they certainly can't apply!
 

FAQ: Finding the Number of Solutions in [0,2pi) for log5(cos x) = -1/2

What is the definition of "Answer in an interval ~ cos"?

"Answer in an interval ~ cos" refers to a mathematical concept where the solution to a problem is given as a range or interval of values, rather than a single numerical answer. The "cos" notation indicates that the problem involves cosine functions.

How is "Answer in an interval ~ cos" different from a regular numerical answer?

Unlike a regular numerical answer, which provides a single value, "Answer in an interval ~ cos" gives a range of values. This allows for a more precise and accurate representation of the solution, as well as accounting for any potential variations or uncertainties in the problem.

When is it necessary to use "Answer in an interval ~ cos" instead of a regular numerical answer?

"Answer in an interval ~ cos" is often used in situations where the problem involves a range of possible values, rather than a single definitive answer. This can include problems with multiple variables, or situations where there may be some degree of uncertainty or variability in the solution.

How can I interpret the "Answer in an interval ~ cos" notation?

The "Answer in an interval ~ cos" notation indicates that the solution to the problem should be given as a range of values, with the "cos" indicating that the problem involves cosine functions. The interval may be given in different forms, such as open or closed intervals, and may also include mathematical symbols such as "less than" or "greater than."

Can "Answer in an interval ~ cos" be used in all types of mathematical problems?

While "Answer in an interval ~ cos" is a commonly used notation, it may not be applicable to all types of mathematical problems. It is most often used in problems involving trigonometric functions, but can also be used in other types of problems where a range of values is more appropriate than a single numerical answer.

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