Answer in an interval ~ cos

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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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Answers and Replies

  • #2
tiny-tim
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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
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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
tiny-tim
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Hi Neophyte! :smile:
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
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Anyway … the answers are 63º and … ? :smile:
:uhh:. 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 :rofl:.

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
tiny-tim
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:uhh:. 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
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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
tiny-tim
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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
HallsofIvy
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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!
 

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