Let S = {x in R : cos(x) + cos(sqrt{2}x)

Question

(D) We are given the set $S = \{x \in R : \cos(x) + \cos(\sqrt{2}x) < 2\}$.We know that the maximum value of the cosine function is $1$,which occurs w

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$S = R - \{0\}$

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(D) We are given the set $S = \{x \in R : \cos(x) + \cos(\sqrt{2}x) We know that the maximum value of the cosine function is $1$,which occurs when the argument is an integer multiple of $2\pi$.For $\cos(x) + \cos(\sqrt{2}x) = 2$,both $\cos(x)$ and $\cos(\sqrt{2}x)$ must simultaneously be equal to $1$.This implies $x = 2n\pi$ and $\sqrt{2}x = 2m\pi$ for some integers $n, m$.If $x = 0$,then $n = 0$ and $m = 0$,which satisfies the condition $\cos(0) + \cos(0) = 1 + 1 = 2$.If $x 
eq 0$,then $\frac{\sqrt{2}x}{x} = \frac{2m\pi}{2n\pi} \implies \sqrt{2} = \frac{m}{n}$,which is impossible since $\sqrt{2}$ is irrational.Thus,the sum $\cos(x) + \cos(\sqrt{2}x)$ is equal to $2$ if and only if $x = 0$,and it is strictly less than $2$ for all $x \in R \setminus \{0\}$.Therefore,$S = R \setminus \{0\}$.

Let $S = \{x \in R : \cos(x) + \cos(\sqrt{2}x) < 2\}$,then

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