Find the range of $a$ such that there exists a line having no intersection with $2\cos x−\cos y=a$

Here's a solution involving roots of unity.

Set $y=2x$, and $c=\cos(x)$. Using double angle identities, we have $$2c-2c^2+1 = -2(c+0.5)^2 + 1.5 = a$$ Hence, no solution for $a>1.5$. Also, $2\cos(x)-\cos(y) \leq |2\cos(x)|+|-\cos(y)| = 2+1 = 3$.

We now claim that $\frac{3}{2} < a \leq 3$. Set $y=mx+b$. For the sake of contradiction, say that there exists some choice of $m,b$ that guarantees a line that does not hit $2\cos(x)-\cos(y)$ for $a \leq \frac{3}{2}$.

From symmetry arguments, we may assume that $m > 0$.

If $m$ is irrational, then as we choose through all periodic $x=2k\pi$, $\cos(y)$ will seem to be "detached" from $\cos(x)$, and the output will be dense in $[-3,3]$. Hence there will exist a value where $\cos(y)<0$, so $2\cos(x)-\cos(y) > 2 > \frac{3}{2}$.

Also taking $x=\pi$ and $2\cos(x)-\cos(y) = -2-\cos(y) < 0 $. Due to continuity and the intermediate value theorem, for any $0< a < \frac{3}{2}$, there will exist an intersection, giving us a contradiction.

Otherwise, $m$ is rational. Hence, set $m=\frac{p}{q}$, where $p,q \in \mathbb{Z}^+$ and $\gcd(p,q)=1$. We need to show that for any choice of $m$, there exists an $x$ such that (after the map $x \mapsto qx$): $$\Re\left(2e^{qxi}\right)-\Re\left(e^{(px+b)i}\right) \geq \frac{3}{2}$$

Remap $x \mapsto \frac{2\pi t}{q}$ in $$\Re\left(2e^{qxi}\right)-\Re\left(e^{pxi+bi}\right) \geq \frac{3}{2}$$ we get $$\Re\left(2e^{\left(2\pi t i\right)}\right) - \Re\left(e^{2\pi \left(\frac{p}{q}\right) ti + bi}\right) \geq \frac{3}{2}$$ and setting $t$ to be integer yields $$\Re\left(e^{2\pi \left(\frac{p}{q}\right) ti +bi}\right) \leq \frac{1}{2}$$ Hence, it suffices to show that there exists some $t$ such that the following does not hold true for some $x,t \in \mathbb{Z}$: $$2k\pi-\frac{\pi}{3} \leq 2\pi \left(\frac{p}{q}\right) t +b \leq 2k\pi+\frac{\pi}{3}$$

Notice we can imagine this to be that $\exp(2\pi \left(\frac{p}{q}\right) ti)$ cannot be in an arc centred around $e^{ib}$ which is $\frac{2\pi}{3}$ wide.

However, for $q>1$, by Bezout's Lemma, there exists a $t$ such that $pt+nq=1$ for some $n \in \mathbb{Z}$, and in fact multiplying by any integer constant $M$ we have $Mpt+Mnq = M$, so in fact there exists $N$ and $t$ where $pt+Nq=M$ for all $M \in \mathbb{Z}$.

Hence, we may now pick $M$ such that $\exp(2\pi \left(\frac{p}{q}\right) ti)$ is no longer in the arc. A similar reasoning for this would be the Pigeonhole Principle as you increment $t$ due to the roots of unity being equally spaced out.

For $q=1$, we note that $\Re(e^{2\pi pti + bi}) = \Re(e^{bi}) \leq \frac{1}{2}$, meaning that we are done unless $\cos(b) > \frac{1}{2}$.

Now, we have that $\cos(b)>\frac{1}{2}$ and $q=1$. Map $x \mapsto \frac{\pi t}{p}$. It suffices to show that

$$\Re\left(2e^{\frac{t}{p} \pi i}\right) - \Re\left(e^{t \pi i + bi}\right) \geq \frac{3}{2}$$

Setting $t=1$ will yield us to need to prove that

$$\Re\left(2e^{\frac{1}{p} \pi i}\right) + \Re\left(e^{bi}\right) \geq \frac{3}{2}$$ and as $\cos(b) > \frac{1}{2}$ this transforms to $$\Re\left(e^{\frac{1}{p} \pi i}\right) = \cos\left(\frac{\pi}{p}\right) \geq \frac{1}{2}$$ which is true for $p \geq 3$.

Hence, we are left with only $p=2,q=1$ or when $m=2$, and when $\cos(b)>\frac{1}{2}$ left.

Taking modulo $2\pi$, if $b \in [0,\frac{\pi}{3})$ then taking $x=\frac{\pi}{3}$ yields

$$2\cos(x)-\cos(y)=2\cos\left(\frac{\pi}{3}\right)-\cos\left(\frac{2 \pi}{3} + b\right) \geq 1 - \frac{-1}{2} = \frac{3}{2}$$ and similarly for $b \in (-\frac{\pi}{3},0]$, then taking $x=-\frac{\pi}{3}$ yields $$2\cos(x)-\cos(y)=2\cos\left(-\frac{\pi}{3}\right)-\cos\left(-\frac{2 \pi}{3} + b\right) \geq 1 - \frac{-1}{2} = \frac{3}{2}$$

Thus, by exhaustion, we are done.

Follow up:

The goal of this answer is to explain why the maxima of the function $f(x,m,b)=2\cos(x)-\cos(mx+b)$, which we designate as $A$, takes the minimum possible value at $b=2k\pi$ where $k \in \mathbb{Z}$.

We may assume that $k \in [-\pi,\pi]$ from periodicity. Say, for the sake of contradiction, the lowest possible minima is achieved at $b=s>0$. Then, from symmetry, minima of all maximums are also achieved at $b=-s$. Graphically, we have that the orange dashed lines do not allow any smaller maxima (indicated here by the dotted green loops).

Now, consider the maximum possible difference in ordinates between the lines (indicated by the black segment). This is achieved at taking the lines $y=\pm \arccos(\frac{1}{2})$, with a maximum difference of $\frac{2\pi}{3}$. However, this difference is way too small to admit an entire orange loop inside of it.

Hence, the entire region between the two dashed orange lines must be empty. Thus, the $y=mx$ line, which fits nicely in it, will not intersect with any of the orange loops.

This means that we may set the minima value $A$ to be lower by some amount (the green loops here) and this will still not intersect $y=mx$. This contradicts the minimality of $y=mx \pm b$. This forms our desired contradiction.

Hence, $b$ cannot be larger than zero. Similarly, it cannot be smaller than zero. Hence, $b=0$ indeed on the interval $[-\pi,\pi]$ and hence $b=2k\pi$ to minimise $A$.

This answer requires the concept of continuity and the Intermediate Value Theorem, so strictly speaking it's a little advanced for high school. But it's hard to talk about intersecting curves without those notions. On the other hand, the use of these notions requires only some graphical intuition that I think is accessible from the high school level.

The proof can be summarized as follows:

First we show that if $a > \frac32,$ the line $y=2x$ satisfies the requirements of the question.

Next we show that if $0<a\leq1,$ only a line parallel to the $y$ axis can avoid intersecting the graph.

Finally we consider $1 < a \leq \frac32.$ Then the graph is a grid of "islands" around the points $(2m\pi,(2n+1)\pi)$ for integers $m,n.$ We show that if the line required by the question exists, there is a line of positive slope passing between the islands around $(0,-\pi)$ and $(0,\pi)$ and not intersecting the graph. We show that this line cannot pass above the island around $(2\pi,3\pi).$ We show that if the line passes below the island around $(2\pi,\pi)$ it lies below the line $y=\pi,$ a contradiction. We show that if the line passes between the islands at $(2\pi,\pi)$ and $(2\pi,3\pi)$ it lies between $y=x+\pi$ and $y=x-\pi$ and has slope $1,$ contradicting the conditions of the question. So no such line exists for $1<a\leq\frac32.$

This leaves us with solutions for $a > \frac32$ and no other positive $a.$

Let $S$ be the set of all positive numbers $a$ such that there is a line $L$ that is neither parallel to an axis nor to $y = x$ such that $L$ does not intersect the graph of $2\cos x - \cos y = a.$

In the question you show the graph for the case $a = \frac32.$ The graph is a grid of "islands." Inside each "island" the value of $2\cos x - \cos y$ is greater than $a$; outside it is less than $a.$

You also propose that the line $y = 2x$ is tangent to the graph. We can parameterize this entire line by its $x$ coordinate so that the value of $2\cos x - \cos y$ at each point is $2\cos x - \cos(2x).$ But $$ 2\cos x - \cos(2x) = 2\cos x - (2\cos^2 x - 1) = -2\left(\cos x-\frac12\right)^2 + \frac32 \leq \frac32. $$ The inequality becomes equality when $\cos x - \frac12 = 0,$ that is, when $x = 2k\pi \pm \frac\pi3$ for an integer $k.$ So the line touches the graph at those values of $x.$ At any other place on the line the value of $2\cos x - \cos y$ is less than $\frac32$ and therefore is in the region outside any of the "islands."

If $a > \frac32$ then the graph of $2\cos x - \cos y = a$ consists only of points strictly inside the "islands" in your graph, and therefore it does not intersect the line $y = 2x.$

So we know at this point that $\left(\frac32, \infty\right) \subseteq S.$ The only question remaining is whether any $\frac32$ or any smaller number can be in $S.$ (The assertion that $S$ is the interval $\left(\frac32, 3\right]$ is incorrect; for $a > 3$ the graph is empty and there is no intersection with any line, including lines that are neither parallel to an axis nor to $y=x$.)

Note that in order for $L$ not to be parallel to the $y$ axis, it must cross the line $x = \pi$ somewhere, and $2\cos x - \cos y \leq 0$ when $x = \pi.$ Since $a > 0,$ this means $L$ must consist entirely of points at which $2\cos x - \cos y < a,$ for if $2\cos x - \cos y = a$ at any point on $L$ then $L$ intersects the graph and if $2\cos x - \cos y = a$ at any point on $L$ then (since $2\cos x - \cos y$ is continuous) it must be true that $2\cos x - \cos y = a$ at some point.

But the line $L$ also must cross the $y$ axis somewhere, and if $a \leq 1,$ then $2\cos x - \cos y \geq a$ at every point on the $y$ axis and no such line $L$ can exist. So $S$ doesn't contain $1$ or any smaller numbers; we only need to consider $1 < a \leq \frac32.$ For such a value of $a,$ we either get the graph in the question (if $a = \frac32$) or we get a similar graph consisting of a grid of "islands," one surrounding each of the "islands" in the graph for $a = \frac32.$ The line $L$ must exist entirely in the region "outside" the "islands," where the value of $2\cos x - \cos y$ is less than $a.$

Note that if we translate the entire diagram upward or downward by an integer multiple of $2\pi,$ the graph is mapped to itself and $L$ is mapped to a parallel line. So it is sufficient to look only at lines that cross the $y$ axis between $y=-\pi$ and $y=\pi$; if any line satisfies the conditions then one of those lines does.

By reflecting across the $y$ axis (if needed) we can make the slope of $L$ be positive. So we need only consider lines with positive slope.

So we assume we have a line $L$ with positive slope passing between the "island" around $(0,-\pi)$ and the "island" around $(0,\pi).$ Remember that these "islands" are either the ones shown in the question (if $a = \frac32$) or larger regions containing those "islands."

The line $L$ must pass above the point $\left(-\frac\pi3, -\frac{2\pi}3\right)$ in or on the island around $(0,-\pi)$ and below the point $\left(\frac\pi3, \frac{2\pi}3\right)$ in or on the island around $(0,\pi).$ That implies the slope of $L$ is less than $2.$ But in order to pass below $\left(\frac\pi3, \frac{2\pi}3\right)$ but above the point $\left(\frac{5\pi}3,\frac{10\pi}3\right)$ on the island around $(2\pi,3\pi),$ $L$ would need to have slope greater than $2.$ So $L$ must pass below the island around $(2\pi,3\pi).$

The line $L$ must cross the line $y = \pi$ somewhere. Moreover it has to do so between two islands. Suppose $L$ crosses $y = \pi$ between the islands around $(2k\pi,\pi)$ and $(2(k + 1)\pi,\pi)$ for some integer $k.$ Then $L$ must pass below the point $\left(2k\pi,\frac{\pi}3\right)$ in or on the first island and the point $\left(2(k + 1)\pi,\frac{5\pi}3\right)$ in or on the other island. So $L$ must have slope greater than $\frac23.$

A line with slope greater than $\frac23$ cannot pass above the point $\left(0,-\frac\pi3\right)$ on the island around $(0,-pi)$ and below the point $\left(2\pi,\frac\pi3\right)$ on the island around $(2\pi,\pi).$ So $L$ has to pass above the island around $(2\pi,\pi).$

So far we have eliminated every possible line $L$ except for lines crossing the $y$ axis between $\left(0,-\frac\pi3\right)$ and $\left(0,\frac\pi3\right)$ and crossing the line $x=2\pi$ between $\left(2\pi,\frac{5\pi}3\right)$ and $\left(2\pi,\frac{7\pi}3\right).$ This restricts the slope of $L$ to be greater than $\frac23$ and less than $\frac43.$ (It actually restricts the slope more tightly than that, but those are bounds that are easy to show using known points in or on the islands.)

Next we show that $L$ cannot cross the line $y=x+\pi$ between $x=2k\pi$ and $x=2(k+1)\pi$ ($k$ an integer) because it would have to pass below $\left(2k\pi,2k\pi+\frac\pi3\right)$ and above $\left(2(k+1)\pi,2k\pi+\frac{5\pi}3\right),$ which would require the slope of $L$ to be greater than $\frac43.$ Similarly we show that $L$ cannot cross the line $y=x-\pi$ because to pass between the islands at $x=2k\pi$ and $x=2(k+1)\pi$ the slope of $L$ would need to be less than $\frac23.$ These proofs are similar to the proof that $L$ cannot cross the line $y=\pi$ while also passing below the island around $(2\pi,\pi),$ so I have omitted some details.

Therefore $L$ must lie between $y=x+\pi$ and $y=x-\pi$ without intersecting either line, so its slope must be $1.$ But $L$ must not be parallel to $y=x,$ so no such line $L$ can exist.

Therefore $a \not\in S$ if $0 < a \leq \frac32.$ We conclude that $S = \left(\frac32,\infty\right).$ The "range" of $a$ requested in the the question is all real numbers greater than $\frac32.$