The function $f(t)=2+\sin(t)+\sin(t\sqrt2)$ The function $f$ defined on $\mathbb{R}$ by
 $$f(t)=2+\sin(t)+\sin(t\sqrt2)$$
 can never reach $0$. Can we find some sequence $(t_n)_{n\geq0}$ such that 
$$\lim_{n \to \infty}f(t_n)=0 \ \ \ ?$$
Or in other words, $\sin(t_n)$ and $\sin(t_n\sqrt2)$ both approaching $-1$.
It seems to me that this is related to how we can approach $\sqrt2$ by rationals.
 A: Consider $t = b \pi/2$ where $b$ is an integer congruent to $3$ mod $4$.  Then $\sin(t) = -1$, so you need $\sin(t \sqrt{2})$ to be near $-1$ as well.  This will be the case if $t \sqrt{2}$ is near $a \pi/2$ for some integer $a$ congruent to $3$ mod $4$.  Thus we want to show that for arbitrary $\epsilon > 0$ there 
are integers $a$, $b$, both congruent to $3$ mod 4, with
$$\left| \dfrac{b \pi}{2} \sqrt{2} - \dfrac{a \pi}{2}\right| < \epsilon$$
i.e.
$$ \left|\sqrt{2} - \dfrac{a}{b}\right| < \dfrac{2 \epsilon}{b\pi}$$
Without the mod 4 restrictions, this would follow easy from Hurwitz's theorem.
With those restrictions, you can use e.g. a result of Tornheim.
EDIT:
Explicitly, it looks to me like you can use
$$ \dfrac{a}{b} = 1 + \cfrac{1}{2+\cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{\ldots + \cfrac{1}{1 + \cfrac{1}{2}}}}}} $$
where the number of 2's in the denominators (including the last one) is congruent to $3$ mod $4$.
EDIT: An easier way is to use the
Equidistribution theorem.  Since $\sqrt{2}$ is irrational, the 
numbers $(n + 3/4) \sqrt{2} \mod 1$ are equidistributed (and, in particular, dense) in $[0,1]$.
