For any $\epsilon>0$, there exists arbitrarily large $x$ with $\cos (x^2)>1-\epsilon$ and $\cos [(x+1)^2]<-1+\epsilon$.
This is an exercise in "Uniform Distribution of Sequences" by Kuipers, Niederreiter.
If I remember correctly, I may have been given a hint that we need to use the fact that the sequence $\left\{\frac {n^2}{2\pi}\right\}_{n=1}^\infty$ is uniformly distributed modulo $1$ (although I'm not sure whether this was indeed the hint), but I can't figure out how to use that.
Also, since this question sounds like a question from Calc I, is there a way to solve this using only properties of the cosine function without using results from Uniform Distribution of sequences?
Note: AlvinL posted this link which does look like the first half of this question, but we should note that this question asks to prove that there are arbitrarily large $x$ which simultaneously satisfy the two given conditions.
Let $x_k=k\sqrt{2\pi}$ for $k\in\mathbb N$, then $\cos(x_k^2) =1$ and $$\cos[(x_k+1) ^2]=\cos(2k^2\pi+2k\sqrt{2\pi}+1) =\cos(2k\sqrt{2\pi}+1).$$
Now, we need AlvinL's link. We have $2\sqrt{2\pi}/(2\pi) $ is irrational. Hence, the fractional part of $( 2k\sqrt{2\pi})/(2\pi)$ is uniformly distributed on $[0, 1]$. Then the fractional part of $( 2k\sqrt{2\pi} +1)/(2\pi)$ is uniformly distributed on $[0, 1]$.
So $(2k\sqrt{2\pi} +1\bmod 2\pi)$ is uniformly distributed in $[0,2\pi]$, which means $\cos( 2k\sqrt{2\pi} +1)$ is dense in $[-1, 1]$. Especially, there is a $k$ with $\cos( 2k\sqrt{2\pi} +1) <-1+\epsilon$.