For what values ​​of a intersects $y = ax$, $y = \sin x$ just one time? As the title says, for what values of a intersects $y = ax$, $y = \sin x$ just one time?

I am not able to solve this problem, and I really want to know the answer.
Sketch $y = \sin x$ and imagine a vertical line through the origin, which initially has only one intersection point at the origin. Notice that we can rotate it clockwise until it is tangent to $y = \sin x$ at some point $x = k_1$, where $k_1 \in (-\pi/2, \pi/2)$. Likewise, notice that we can rotate it counterclockwise until it is tangent to $y = \sin x$ at some point $x = k_2$, where $k_2 \in (-2\pi, -\pi)$.

To solve for $k_1$ and $k_2$, we need to know when the tangent line of $y = \sin x$ at $x = k$ has a slope that is equal to the slope of the line passing through the points $(k, \sin k)$ and $(0, 0)$. That is, we must solve:
$$
\cos k = \frac{\sin k - 0}{k - 0} \iff \tan k = k
$$
Solving this equation with Wolfram|Alpha, we see that $k_1 = 0$ and $k_2 = 4.4934\ldots$. Taking their slopes, we have $\cos k_1 = 1$ and $\cos k_2 = -0.2172\ldots$ so that:
$$
a \in (-\infty, -0.2172) \cup (1, \infty)
$$
One intercept is at $x=0$.
Next, $ax=\sin x$, taking derivative of both sides wrt $x$, we get $a=cos x$.
Then, $x=\cos^{-1} a$ is another intercept. So $x=0$ will be the only intercept if $\cos^{-1}a$ does not exist, or $\cos^{-1}a=  0$, which will be the case if $a&gt;=1$ or $a&lt;-1$.