# Answer in an interval ~ cos

Neophyte

## 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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Homework Helper
Hi Neophyte! (have a pi: π and a degree: º and a square-root: √ and try using the X2 tag just above the Reply box )

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? Neophyte
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.

Homework Helper
Hi Neophyte! 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? )

Sorry, but I don't understand the rest of your post. Anyway … the answers are 63º and … ? Neophyte
Anyway … the answers are 63º and … ? :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; 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. But (und) was suppose to be undefined probably inaccurate though.

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Homework Helper
: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 … ? Neophyte
Yes, but didn't the question say [0,2pi)?

Nooo … sine is a different shape graph …

sin22º = sin … ? 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 Do the arc limitations even play a role in this?

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Homework Helper
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! Doh, So it would be 22º and 158º ? But that is not in the interval for arcsin 