# Find where cos(x) = acos(x)

• irishetalon00
In summary, to find where cos(x) = acos(x), graph y = cos(x) and y = acos(x) and find the point of intersection. Alternatively, you can solve the equation x = cos(cos(x)) and substitute the solution into the original equation to check for consistency.

## Homework Statement

Find where cos(x) = acos(x)

Trig identities?

## The Attempt at a Solution

I'm not sure how to proceed with this analytically. Cos(cos(x))=x? that doesn't seem to help...

irishetalon00 said:

## Homework Statement

Find where cos(x) = acos(x)

Trig identities?

## The Attempt at a Solution

I'm not sure how to proceed with this analytically. Cos(cos(x))=x? that doesn't seem to help...
Think about the period and phase relationships between y1 = cos (x) and y2 = a cos (x).

Where are the only values of x located where each function y1 and y2 could possibly intersect?

Hint: It's OK to make a sketch of these two functions.

irishetalon00 said:

## Homework Statement

Find where cos(x) = acos(x)
Does acos(x) mean a * cos(x) or arccos(x)?
irishetalon00 said:

Trig identities?

## The Attempt at a Solution

I'm not sure how to proceed with this analytically. Cos(cos(x))=x? that doesn't seem to help...

I presume the acos(x) is arccos(x). The x would be measured in radians when taking the cosine. Your graphical solution and or tabulate the two as a function of x might work best. I think you will find the solution near ## x=\pi/4 ## for which ## \cos(\pi/4)=.7071... ##. On the other hand if "a" is simply a constant, see @SteamKing post #2. editing...besides a tabulation, Taylor series of the two functions (near the point(s) above) can provide for a more analytical solution.

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I presume the acos(x) is arccos(x).
As do I, based on what the OP wrote about cos(cos(x)) in post 1. Still, a question should be posted so that it isn't ambiguous. acos(x) is notation that is used in programming, and not as much in a mathematics context.

Upon solving this one, (see post #4), it appears the problem could also be stated as find x (angle in radians) such that x=cosx, because x=cos(x)=arccos(x) is a direct consequence of the first equality. Thereby, you just need to graph y=x and y=cos(x) and find the point of intersection.

SammyS
Upon solving this one, (see post #4), it appears the problem could also be stated as find x (angle in radians) such that x=cosx, because x=cos(x)=arccos(x) is a direct consequence of the first equality. Thereby, you just need to graph y=x and y=cos(x) and find the point of intersection.

No. The correct equation would be ##x = \cos( \cos x))##.

Ray Vickson said:
No. The correct equation would be ##x = \cos( \cos x))##.
@Ray Vickson Please look more closely at my post #6. I think I have this one correct. If x=cos(x) then (taking arccos of both sides) arccos(x)=x and thereby cos(x)=arccos(x). I think the converse is true for this case: if cos(x)=arccos(x) I think x=cos(x). .. editing...if you take x=cos(cos(x)) and suppose a solution of cos(x)=x then it reads (after substituting in the inner term) x=cos(x) and you have consistency.

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@Ray Vickson Please look more closely at my post #6. I think I have this one correct. If x=cos(x) then (taking arccos of both sides) arccos(x)=x and thereby cos(x)=arccos(x). I think the converse is true for this case: if cos(x)=arccos(x) I think x=cos(x). .. editing...if you take x=cos(cos(x)) and suppose a solution of cos(x)=x then it reads (after substituting in the inner term) x=cos(x) and you have consistency.

Take the ##\cos## on both sides of ##\cos(x) = \arccos(x)## and you get ##x = \cos(\cos x))##. Apparently, we must have ##\cos x = \cos( \cos x)## at the solution. So, your way is correct, but so is the alternative.

Ray Vickson said:
Take the ##\cos## on both sides of ##\cos(x) = \arccos(x)## and you get ##x = \cos(\cos x))##. Apparently, we must have ##\cos x = \cos( \cos x)## at the solution. So, your way is correct, but so is the alternative.
The graphical solution of cos(x)=arccos(x) is interesting. Graphing y=cos(x) and y=arccos(x) and finding where they intersect. The second graph is x=cos(y) so they have a symmetry to them and will meet along y=x, which is the result from the alternative method (post #6). We still need to hear from the OP...

## 1. What does cos(x) = acos(x) mean?

This equation means that the cosine of a given angle (x) is equal to the inverse cosine of that same angle. In other words, the angle (x) is the same for both the cosine and inverse cosine functions.

## 2. How can I solve for x in the equation cos(x) = acos(x)?

To solve for x, you can use a trigonometric identity known as the inverse cosine function (arccos). This function will give you the value of x that satisfies the equation cos(x) = acos(x).

## 3. Can there be more than one solution to cos(x) = acos(x)?

Yes, there can be multiple solutions to this equation. Since the cosine function is periodic, it will repeat itself after every 2π radians (or 360 degrees). This means that there can be infinitely many angles that satisfy cos(x) = acos(x).

## 4. How can I graph the solutions to cos(x) = acos(x)?

You can graph the solutions by plotting the points where the cosine and inverse cosine functions intersect on a coordinate plane. These points will form a straight line at the angle x.

## 5. Are there any real-world applications for solving cos(x) = acos(x)?

Yes, this equation can be used in various fields such as engineering, physics, and astronomy. For example, it can be used to calculate the phase difference between two oscillating systems or the angle of incidence in optics.

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