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Let $a \in \mathbb R$, what values of $t$ solve the equation $at + \sin(t) = 0$?

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First, $t=0$ is a solution. Otherwise, it can be written as $$-a=\frac{\sin t}t.$$ One can find the bounds of $\frac{\sin t}t$ so that if $-a$ is out of that range, there is no solution. But, in general, it seems only numerical solutions could be found, see also its graph.

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Since we can write it as $at=-\sin(t)$, the roots lie where the line with gradient $a$ and $-\sin(t)$ intersect.
If $a\le1$, the only root is $0$.
For $-1<a<0$ there are finitely many solutions whose values can only be computed numerically. There will be no roots when $t>T$, where $T$ is such that $|aT|>1$ (i.e. $T>1/|a|$ or $T<-1/|a|$).
The closer $a$ is to $0$, the more roots there are. At $a=0$, there are infinitely many roots at $t=n\pi$. For all $n \in \mathbb{Z}$.
If $0<a\le A$, there are again finitely many solutions. ($A$ is a constant such that $a=A$ has exactly 1 solution)
When $a>A$, there are zero solutions.
Notice that when $a=A$, the line is tangent to the curve, meaning that for some $t_{0}$, $At_{0}=-\sin(t_{0})$ and $A=-\cos(t_{0})$. Dividing these, we get $t_{0}=\tan(t_{0})$. We can solve numerically to get $t_{0}=4.493409...$ and thus $A=0.217234...$

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The function $$\operatorname{sinc}(x)\equiv\frac{\sin(x)}{x}$$ is called the $\operatorname{sinc}$ function. You need the (multi-valued) inverse of this function, which is a transcendental function with no name. We could call it the "arcsinc" function. Defining $\operatorname{arcsinc(x)}$ by $$ \operatorname{sinc}(\operatorname{arcsinc(x)})\equiv1,$$ the solution to your equation is $$t=\operatorname{arcsinc(-a)}.$$

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