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

[bsmithysmith]

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.

 

[DavidCampen]

http://mathhelpboards.com/rules/

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.

 

[Nono713]

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 $2\pi/4$).

 

[bsmithysmith]

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.

 

[DavidCampen]

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.

 

[Prove It]

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 $\displaystyle \begin{align*} 2\pi \end{align*}$.

 

[DavidCampen]

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.