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This answer requires the concept of continuity and the Intermediate Value Theorem, so it's a little advanced for high school. But it's hard to talk about intersecting curves without those notions.

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.

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 $2\cos x - \cos y = a.$

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 $2\cos x - \cos y = a.$

Next we need to 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 need to 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 $\frac23.$ 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. These proofs are similar to the proof that $L$ cannot cross the line $y=\pi$ if its slope is less than $\frac13,$ so I have omitted some details.

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.