Why do we need to find the intersection between these lines?

Question

We have the functions

$$ x = -1 + 2 \cos(t)$$

$$ y = 3 + 2 \sin(t)$$

They give P's orbit

with $t$ on $\left[0, \dfrac{3}{2} \pi\right]$

Find (to 2 decimal places accurate) for which values of t the Point t is left of the line $x=-2.$

So we're allowed to use our calculators, and the answer model says that to find the solution you first need to find the intersection between $\;y=-2\;$ and $\;y=-1 + 2\cos (x)$.

My question is; why does this work?

Answer 1

I agree with Ross Millikan that there is a typo, and the suggestion is to solve the intersection between $x = -2$ and $x = -1 + 2\cos t$.

Solving for $t$ at that intersection will give you the value of $t$ exactly where $x = -2$. Knowing that, you can then determine which values of $t$ must therefore lie to the left of the line $x = -2$.

Essentially, the lines will intersect when $x = -2$ and $x = -1 + 2\cos t$, i.e., when $$-2 = -1 + 2\cos t\iff 2\cos t = -1\iff \cos t = \frac{-1}{2}$$

So, when $\cos t = -\frac 12$, we're on the line $x = -2$. So when $\cos t \lt -\frac 12$, we're to the left of the line $x = -2$.

Answer 2

I don't know. To be left of $x=-2$ you need $\cos t \lt -\frac 12$ You should be able to find appropriate $t$ without worrying about $y$. I think they are finding the intersection between $x=-2$ and $x=-1+2 \cos t$ and the $y$ is a typo.