Finding $a\in\mathbb{R}$ for which the curve $\gamma(x)=\left(x+a\sin(x), a\cos(x) \right)$ is self-intersecting

Question

How to find the $a\in\mathbb{R}$ for which the curve $$\gamma(x)=\left(x+a\sin(x), a\cos(x) \right)$$ is self-intersecting?

Is it possible to calculate analytically?

Answer 1

Let $a\neq 0$, $a \cos x_1=a \cos x_2 \Rightarrow$ $x_2=x_1+2k\pi \lor x_2=-x_1+2k\pi$

$x_2=x_1+2k\pi, x_2+a\sin x_2=x_1+a\sin x_1 \Rightarrow x_2=x_1$, not self-intersecting when $x_2-x_1=2k\pi$.

$x_2=-x_1+2k\pi, x_2+a\sin x_2=x_1+a\sin x_1 \Rightarrow 2a \sin x_1=2k\pi-2x_1$

$x_1=n\pi \Rightarrow \sin x_1=0 \Rightarrow 2k\pi-2x_1=0 \Rightarrow x_1=k\pi\Rightarrow x_2=-x_1+2k\pi=k\pi=x_1$, not self-intersecting when $x_1=n\pi$.

$x_1\neq n\pi \Rightarrow \sin x_1\neq 0 \Rightarrow a=\frac{k\pi-x_1}{\sin x_1}$. Maximum negative $a$ is $-1$, obtained when $k=2n$, $x_1\to k\pi$, minimum positive value $a$ is $1$, obtained when $k=2n+1$, $x_1\to k\pi$.

Limiting values $a=\pm 1$ don't give self-intersection because they are only attainable as the limit at $x_1\to k\pi$ and $x_1=k\pi$ is not self-intersection point.

Therefore, the curve is self-intesecting when $|a|>1$.