5cos(4x)=4 how to find the third smallest solution?

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Discussion Overview

The discussion centers around solving the equation $$5\cos(4x)=4$$ for its third and fourth smallest solutions. Participants explore methods for finding solutions, the cyclical nature of the cosine function, and the implications of the equation's periodicity.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant describes their approach to finding the first two solutions using inverse cosine and division by 4, but expresses uncertainty about finding the third solution.
  • Another participant suggests that there are infinitely many solutions and emphasizes the need to clarify which specific solutions are being sought.
  • A later reply indicates that the instructor expects students to understand the cyclical nature of the cosine function and suggests adding multiples of $$2\pi$$ to the primary solutions to find additional solutions.
  • One participant corrects a previous statement by noting that the period of the function is actually $$\frac{\pi}{2}$$ due to the factor of 4 inside the cosine function, rather than $$2\pi$$.
  • Another participant acknowledges their oversight in not calculating the actual answers and recognizes the importance of the factor of 4 in the cosine function.

Areas of Agreement / Disagreement

Participants generally agree on the cyclical nature of the cosine function and the method of generating additional solutions, but there is some disagreement regarding the specifics of the periodicity and the exact solutions being sought.

Contextual Notes

There is a lack of clarity regarding the specific values of the first two solutions and how they relate to the third and fourth solutions. The discussion also reflects varying interpretations of the periodicity of the function based on the equation's structure.

bsmithysmith
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I am given to solve:

$$5\cos(4x)=4$$

I know how to find the first and second solutions, it's basically the inverse, then divide by the 4. The second on is the same process, but $$2\pi$$- the cosine inverse. But I don't quite understand how to get the third solution.
 
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10. Do not post part of the question in the post title and the rest in the main body of the post. Post all of the question in the main body of the post.

It would be helpful if you would post the answers that you do know.
 
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There are infinitely many solutions to this equation, so you might want to clarify which ones you are asked to find. I imagine you need to find the two principal ones, in other words, the two solutions $x_1$ and $x_2$ "closest" to zero. All the other solutions are then just multiples of $\frac{2\pi}{4}$ away.

So, what exactly is the third solution? Looks to me like you found them all (up to multiples of $2\pi/4$).
 
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The two of them were the easiest to find, I didn't clarify my question exactly, too. I'm trying to find the third and fourth smallest solution since I already found two.
 
The instructor wants you to demonstrate that you understand that the cosine function is cyclical and repeats every $$2\pi$$. You have the 2 primary answers that are in the 1st and 4th quadrants. Now just continue adding multiples of $$2\pi$$ to those answers to generate as many additional answers as you want.
 
DavidCampen said:
The instructor wants you to demonstrate that you understand that the cosine function is cyclical and repeats every $$2\pi$$. You have the 2 primary answers that are in the 1st and 4th quadrants. Now just continue adding multiples of $$2\pi$$ to those answers to generate as many additional answers as you want.

Except in this case, the period of the function is $\displaystyle \begin{align*} \frac{2\pi}{4} = \frac{\pi}{2} \end{align*}$, NOT $\displaystyle \begin{align*} 2\pi \end{align*}$.
 
Sigh, yes, I should have calculated the actual answers instead of just eyeballing the question but even so I should have noticed the factor of 4 inside the cosine function.
 

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