How to solve this inequation? $\left|x\sin5x\right|<\frac{\sqrt{2}}{2}$ Im looking for all $x\geq0$ that satisfy $f(x)\colon=\left|x\sin5x\right|<\frac{\sqrt{2}}{2}$.
What I know is that for each $x_{k}=\frac{\pi k}{5}$ we get that $f(x_{k})=0$.
Also, $f(x)$ is a continuous function, so for each of those $x_{k}$'s
there is a $\delta_{k}>0$ such that for all $x\in\left(x_{k}-\delta_{k},x_{k}+\delta_{k}\right)$
we get that $f(x)<\frac{\sqrt{2}}{2}$. Further I suppose that as $x$ gets
bigger, those $\delta_{k}$'s have to get smaller.
My question is
what $\delta_{k}$'s are enough to satisfy that? Can we find the biggest $\delta_{k}$'s ?
Edit 
That is to say im looking to represent $\delta_{k}$
as a function of $k$. 
Would it be easier to solve $\left|x\sin5x\right|<1$
instead?
I think for example that for $x_{k}=\frac{\pi k}{5}$ and for $\delta_{k}=\frac{1}{5\left(x_{k}+1\right)}=\frac{1}{\pi k+5}$
we get that all $x\in\left[x_{k}-\delta_{k},x_{k}+\delta_{k}\right]$
are good enough to solve it, though I have no idea how
to prove that. Any thoughts?
 A: Rewrite $\sqrt{2}|\sin 5x| \lt \frac{1}{|x|}$ and plot each side of the inequality. What do you see?
As we see from the below figure, the equation $\sqrt{2}|\sin 5x| = \frac{1}{|x|}$ has infinitely many positive roots. Let us denote the set of these roots by $R=\{r_1,r_2,r_3,\dots\}$. From the following plot the set of all $x$ satisfying the inequality is
$$S=[0,r_1)\cup(r_2,r_3)\cup(r_4,r_5)\cup\dots$$
but unfortunately the roots of that equation cannot be found in closed form so this is the most we can get. Also, note that for $x_k=\frac{\pi k}{5}$ we cannot have $x_k=\frac{r_n+r_{n+1}}{2}$ for some $n$ and consequently the intervals of the form $(x_k-\delta_k,x_k+\delta_k)$ does not contain all of the $x$'s satisfying the inequality or contain some $x$'s which does not satisfy the inquality.

A: The endpoints of your intervals are solutions of $x \sin(5x) = \pm \sqrt{2}/2$.
These will not be expressible in closed form, so you can't explicitly "find the biggest $\delta_k$'s".  However, you can estimate them: if $0 < a < x < b$ then $\sin(5x) = \pm \frac{\sqrt{2}}{2x}$ is between $\pm\sqrt{2}/(2a)$ and $\pm\sqrt{2}/(2b)$. 
A: Apperantly, this holds for what i was looking for:
Lets look for $x\geq0$ (Not All) that satisfy $f(x)\colon=\left|x\sin5x\right|<1$.
Clearly, for each $k\in\mathbb{N}$ and $x_{k}=\frac{\pi k}{5}$ we
get that $f(x_{k})=0$, so those already solve our inequation. Also,
$f(x)$ is continuous so as we know, each $x_{k}$ has a $\delta_{k}>0$
such that for all $x\in\left(x_{k}-\delta_{k},x_{k}+\delta_{k}\right)$
we get that $f(x)<1$. Now instead of those $\delta_{k}$'s, we will
find $\delta_{k}>\alpha_{k}>0$ such that all $x\in\left[x_{k}-\alpha_{k},x_{k}+\alpha_{k}\right]$
satisfy what we need, and that is enough for what i was looking for.
We will begin by looking for non negative $t_{k}$'s that satisfy
$$
\left|\left(x_{k}+t_{k}\right)\sin5\left(x_{k}+t_{k}\right)\right|<1
$$
With the help of the inequation $\left|\sin x\right|\leq\left|x\right|$
we get
$$
\begin{aligned} & \left|\left(x_{k}+t_{k}\right)\sin5\left(x_{k}+t_{k}\right)\right|=\left(\frac{\pi k}{5}+t_{k}\right)\left|\sin5\left(\frac{\pi k}{5}+t_{k}\right)\right|=\\
 & \qquad=\left(\frac{\pi k}{5}+t_{k}\right)\left|\sin\left(\pi k+5t_{k}\right)\right|=\left(\frac{\pi k}{5}+t_{k}\right)\left|\sin5t_{k}\right|\leq\\
 & \qquad\leq\left(\frac{\pi k}{5}+t_{k}\right)\cdot5t_{k}
\end{aligned}
$$
Therefore, it is enough to find solutions for $\left(\frac{\pi k}{5}+t_{k}\right)\cdot5t_{k}<1$
or for
$$
5t_{k}^{2}+\pi kt_{k}-1<0
$$
The solutions for $5t_{k}^{2}+\pi kt_{k}-1=0$ are
$$
t_{1_{k}},t_{2_{k}}=\frac{-\pi k\pm\sqrt{\left(\pi k\right)^{2}+20}}{10}
$$
And lets denote $t_{1_{k}}<0$ and $t_{2_{k}}>0$. 
Another thing to notice is that
$$
\begin{aligned} & \sqrt{\left(\pi k\right)^{2}+20}+\pi k<\sqrt{\left(\pi k\right)^{2}+4\pi^{2}}+\pi k=\pi\sqrt{k^{2}+4}+\pi k=\\
 & \qquad=\pi\left(\sqrt{k^{2}+4}+k\right)\leq\pi\left(\sqrt{k^{2}+4k+4}+k\right)=2\pi\left(k+1\right)
\end{aligned}
$$
Now from all of that we get that
$$
\begin{aligned} & t_{2_{k}}=\frac{\sqrt{\left(\pi k\right)^{2}+20}-\pi k}{10}=\frac{\sqrt{\left(\pi k\right)^{2}+20}-\pi k}{10}\cdot\frac{\sqrt{\left(\pi k\right)^{2}+20}+\pi k}{\sqrt{\left(\pi k\right)^{2}+20}+\pi k}=\\
 & \qquad=\frac{2}{\sqrt{\left(\pi k\right)^{2}+20}+\pi k}>\frac{2}{2\pi\left(k+1\right)}=\frac{1}{\pi\left(k+1\right)}=\colon\alpha_{k}>0
\end{aligned}
$$
So finally we can see that all $x\in\left[\frac{\pi k}{5}-\frac{1}{\pi(k+1)},\frac{\pi k}{5}+\frac{1}{\pi(k+1)}\right]$
satisfy $\left|x\sin5x\right|<1$.
