When do the combined heights of two Ferris wheels reach 27.5m?

[Amathproblem22]

So I have figured out two equations model a Ferris wheel ride(two different ferris wheels).
F1 = Ferris wheel one
F2= Ferris wheel two

F1, \(\displaystyle h=-12\cos\frac{\pi}{10}t+12.5\)

F2, \(\displaystyle h=-12\cos\frac{\pi}{30}t+15\)

Now from these two equations, I want to know when F2+F1= 27.5m i.e I want to find the times when the combined height of each Ferris wheels seat adds up to 27.5m.
How would I go about doing this?

 

[MarkFL]

If we equate sum of the two height functions to 27.5, we ultimately obtain:

\(\displaystyle \cos\left(\frac{\pi}{10}t\right)+\cos\left(\frac{\pi}{30}t\right)=0\)

Let's let:

\(\displaystyle \theta=\frac{\pi}{30}t\)

And we may write:

\(\displaystyle \cos(3\theta)+\cos(\theta)=0\)

Applying a triple-angle identity for cosine, we ultimately get:

\(\displaystyle 2\cos^3(\theta)-\cos(\theta)=0\)

Factor:

\(\displaystyle \cos(\theta)(2\cos^2(\theta)-1)=0\)

Can you proceed?

 

[Amathproblem22]

If we equate sum of the two height functions to 27.5, we ultimately obtain:

\(\displaystyle \cos\left(\frac{\pi}{10}t\right)+\cos\left(\frac{\pi}{30}t\right)=0\)

Let's let:

\(\displaystyle \theta=\frac{\pi}{30}t\)

And we may write:

\(\displaystyle \cos(3\theta)+\cos(\theta)=0\)

Applying a triple-angle identity for cosine, we ultimately get:

\(\displaystyle 2\cos^3(\theta)-\cos(\theta)=0\)

Factor:

\(\displaystyle \cos(\theta)(2\cos^2(\theta)-1)=0\)

Can you proceed?

\[ \cos \left(θ\right)=0\quad \mathrm{or}\quad \:2\cos ^2\left(θ\right)-1=0 \]

\[ \cos \left(θ\right)=0: θ=\frac{\pi }{2}+2\pi n,\:θ=\frac{3\pi }{2}+2\pi n \]

\[ 2\cos ^2\left(θ\right)-1=0: θ=\arccos \left(\sqrt{\frac{1}{2}}\right)+2\pi n,\:θ=2\pi -\arccos \left(\sqrt{\frac{1}{2}}\right)+2\pi n,\:θ=\arccos \left(-\sqrt{\frac{1}{2}}\right)+2\pi n,\:θ=-\arccos \left(-\sqrt{\frac{1}{2}}\right)+2\pi n \] ?

 

[MarkFL]

Yes, for:

\(\displaystyle \cos(\theta)=0\)

This implies (where \(k\in\mathbb{Z}\)):

\(\displaystyle \theta=\frac{\pi}{2}+\pi k=\frac{\pi}{2}(2k+1)\)

\(\displaystyle t=\frac{30}{\pi}\theta=15(2k+1)\)

And for:

\(\displaystyle \cos(\theta)=\pm\frac{1}{\sqrt{2}}\)

This implies:

\(\displaystyle \theta=\frac{\pi}{4}+\frac{\pi}{2}k=\frac{\pi}{4}(2k+1)\)

\(\displaystyle t=\frac{30}{\pi}\theta=\frac{15}{2}(2k+1)\)