Solve an inequality

Question

How do I solve this inequality for x? How can I solve for x in these kinds of equations in general?

$$\left|\frac{\sin(x)}{x}\right| < \varepsilon$$

Answer 1

In general, you can't do it exactly.

I will set the absolute values aside. We will soon see why they don't really matter for your question. It would just be a per case distinction to solve $\frac{\sin(x)}{x}<\varepsilon$ and $\frac{-\sin(x)}{x}<\varepsilon$ anyway. Also for the moment let's consider equality rather than inequality because the equal case is the border case for the inequality solutions and usually helps finding them very much.

We have $\sin(x)$ = $\frac{1}{2i}(e^{ix}-e^{-ix})$, we need to solve $$\frac{1}{2i}\left(e^{ix}-e^{-ix}\right) = x\varepsilon$$ Behind this lies the problem of solving equations like $z=ae^z$, but we can't do this exactly. There is something called the Lambert W-Function $W$ which gives us the (or better: a) inverse for the function $f(z)=ze^z$. $W$ cannot be expressed in "basic" mathematical terms, we can't write it down with finite terms, not even as something complicated like $e^{\sin(\log z)}z^{\frac{1}{z}}$. We can only get its values numerically. So, we can express the solutions of $ae^z$ in terms of the $W$ Function, it would be $-W(-a)$ (found here, assuming $a\neq 0$). If there is any way to solve $\sin(x)=x\varepsilon$, it would probably like that. Wolfram Alpha doesn't have enough calculation power to do that for me, but it gives a nice graph over the solution space, see here.

See also Roots of $f(x)=\sin(x)-ax$

Answer 2

Note that

$$\left|\frac{\sin x}{x}\right|<1$$

thus the inequality

$$\left|\frac{\sin x}{x}\right|<k$$

is always true $\forall x \neq 0$ when $k\ge 1$.

Otherwise we can have 2 or more solution depending on the value of k.

Plot of $\sin x/x$

Answer 3

The equation associated to the inequation cannot be solved analytically an requires numerical methods. Anyway, you cannot dispense yourself of root isolation.

From the plot of $|\text{sinc } x|$, it is obvious that there are two roots per interval $[k\pi,(k+1)\pi]$ where the maximum reaches $\epsilon$. The maxima are realized at the roots of the derivative, which are solutions of

$$x=\tan x,$$ also to be found numerically. They are $0$, and numbers closer and closer to the vertical asymptotes of the tangent function, at $\dfrac{k\pi}2$.

Let $x_k$ be the locations of these maxima. Then for all $k$ such that $$\left|\dfrac {\sin x_k}{x_k}\right|>\epsilon,$$ we have two roots at some $x'_k$ and $x''_k$, comprised between successive multiples of $\pi$.

Then the solution of the inequation is made of open intervals $(x'_k,x''_k)$, and with $K$ being the last $k$ for which the equation has a solution, $[x'_K,\infty)$ and symmetrically in the negatives.

Finally, note that when $\epsilon>1$, the largest maximum is exceeded and the solution set is $\mathbb R$.