$\{x\in\mathbb R:x\sin x\le 1,x\cos x\le 1\}$ closed or open $?$ Is the set $\left\{x\in\mathbb R:x\sin x\le 1,x\cos x\le 1\right\}$ closed or open or bounded$?$
$$\left\{x\in\mathbb R:x\sin x\le 1,x\cos x\le 1\right\}=\left\{x\in \mathbb R:\sin x\le {1\over x},\cos x\le {1\over x}\right\}$$
Now,whatever the $x$ be , $\sin x\le 1$ and $\cos x\le 1$  always hold.
If $\sin x\gt 0$ then we need $x\le 1$ so that $x\sin x\le 1$ is true. When $\sin x\le 0$ then any  finite value of $x$ is accepted.Same for $\cos x$. So the set we have is basically 
$\{x\in \mathbb R: \sin x\gt0,\cos x\gt 0 ,x\le 1\}\cup \{x\in \mathbb R:\sin x\le 0,\cos x\gt 0 , x\le 1\}\cup \{x\in \mathbb R: \cos x\le0,\sin x\le 0\}\cup \{x\in \mathbb R:\cos x\le 0,\sin x \gt 0 , x\le 1\}$   
So , if I can find these four sets,I'll know the answers. But what are the real number equivalents of the radian angles $\pi,{\pi\over 2},{\pi\over 3}...$ etc?
Or may be I should go for the series expansion of $\sin$ and $\cos.$
$$\sin(x)= x-{1\over 3!}x^3+{1\over 5!}x^5-{1\over 7!}x^7+..$$
$$\cos(x)= 1-{1\over 2!}x^2+{1\over 4!}x^4-{1\over 6!}x^6+.. $$and try to find out for what $x$ these two take what values? I can may be do that with taking just the first two or three terms of the series. But I think there has to be a more useful method or a trick that solves this. 
Please I need some guidance. Thanks. 
 A: Take $f,g : \mathbb{R} \to \mathbb{R}$ be the functions
$$
f(x)= x\sin(x) \text{ and } g(x) = x\cos(x)
$$
Then the set in question is
$$
S = \{x\in \mathbb{R} : f(x)\leq 1, g(x) \leq 1\}
$$
Since $f$ and $g$ are continuous, if $(x_n) \subset S$ such that $x_n \to x$, then
$$
f(x_n) \to f(x) \text{ and } g(x_n) \to g(x)
$$
Since $f(x_n) \leq 1$ for all $n\in \mathbb{N}$, it must follow (why?) that $f(x) \leq 1$. Similarly, $g(x) \leq 1$, and so
$$
x\in S
$$
Edit: As indicated by a commenter, I include an explanation as to why this implies that $S$ contains all its limit points: If $x\in \mathbb{R}$ is a limit point of $S$, then for each $n \in \mathbb{N}$, the open set
$$
(x-1/n,x+1/n)
$$
must intersect $S$ non-trivially. Choose a point $x_n \in (x-1/n,x+1/n)\cap S$. Now note that $|x_n - x| < 1/n$, so $x_n \to x$. Now complete the argument as above to conclude that $x\in S$.
A: 1.There are many equivalent definitions of continuity.One is that $f^{-1}Y$ is closed in the domain whenever $Y$ is closed in the range. 
2.The product of two continuous real functions is continuous. And $x, \sin x,$ and $\cos x$ are continuous functions of $x.$ Therefore $f(x)=x\sin x$ and $g(x)=x\cos x$ are continuous.


*$S=\{x:f(x)\leq 1\land g(x)\leq 1\}=(\;f^{-1}(-\infty,1]\;)\cap (\;g^{-1}(-\infty,1]\;)$ is the intersection of two closed sets so $S$ is closed. 

*$S$ is not empty, as $\pi\in S.$ And $S\ne \mathbb R,$ as $\pi/2\not \in S$ (because $f(\pi /2)=\pi /2>1).$ The only subsets of $\mathbb R$ that are both open and closed are  $\mathbb R$ and $\emptyset.$ So $S$ is not open.

*$S$ is not bounded because  $n\pi\in S$ for every odd $n\in \mathbb N.$
6.(Footnote). $S\ne \{x\in \mathbb R: \sin x\leq 1/x \land \cos x\leq 1/x\}$ for two reasons: (i). $\;0\in S.$ (ii). For $x<0$ we have $(x\sin x\leq 1\land x\cos x\leq 1)\iff (\sin x\geq 1/x\land \cos x\geq 1/x).$ For example  $(-2\pi)\in S.$
