Show that the absolute of the shifted sine and the scaled sine coincide twice Let $\alpha\in(0,\pi)$ and $c>0$. How do I show that there exist exactly two $x\in[0,\pi]$ with $|\sin(x-\alpha)|=c\sin(x)$?
Let $f(x)=|\sin(x-\alpha)|$ and $g(x)=c\sin(x)$. Let $\alpha\le\pi/2$. We know that $f(0)>0=g(0)$, that $f$ is strictly decreasing on $[0,\alpha]$ while $g$ is strictly increasing, so there exists exactly one $x\in(0,\alpha)$ such that $f(x)=g(x)$. On $[\alpha,\pi]$ it's tricky because the derivatives $f'(x)=\cos(x-\alpha)$ and $g'(x)=c\cos(x)$ initially and finally have the same signs, and the derivatives are not simpler compared to the original problem. The case $\alpha\in[\pi/2,\pi)$ is then immediate by looking at $f(\pi-x)$ and $g(\pi-x)=g(x)$.
While this question is related, the answer is to specific to apply in this case.
Update: I want to stress that I in particular ask for a proof that there is no more than one intersection point on $(\alpha,\pi)$.

By using the Existence Theorem of Zero Points, we can prove there exists at least one intersection point on $(\alpha,\pi)$.
Before starting, I'd define $W=f(x)-g(x)$ for $x\in[\alpha,\pi+\alpha]$. Notice that
\begin{align*}
W&=\sin(x-\alpha)-c\sin(x)\Rightarrow W=\sin(x)\cos(\alpha)-\cos(x)\sin(\alpha)-c\sin(x)\\
&=(\cos(\alpha)-c)\sin(x)-\cos(x)\sin(\alpha).
\end{align*}
①$f(\frac{\pi}{2})>g(\frac{\pi}{2})$, the intersection point is on $(\alpha,\frac{\pi}{2})$, $c<\sin(\frac{\pi}{2}-\alpha)$ , that is $\cos(\alpha)>c$.
On $[\alpha,\frac{\pi}{2}]$, the range of $f(x)$ includes the range of $g(x)$, and two funcions are monotonically increasing. There has one and only one intersection point on $[\alpha,\frac{\pi}{2}]$.
Consider $(\frac{\pi}{2},\pi)$. Since $\cos(\alpha)>c$ , $\cos(x)<0$, $W>0$ holds good on $(\frac{\pi}{2},\pi)$. Therefore, two functions do not have any intersection point on $(\frac{\pi}{2},\pi)$.
②$f(\frac{\pi}{2})<g(\frac{\pi}{2})$ , the intersection point is on ($\frac{\pi}{2},\pi$).
(i) $f(\alpha+\frac{\pi}{2})<g(\alpha+\frac{\pi}{2})$ . According to the monotonicity， there exists only one intersection point on $(a+\frac{\pi}{2},\pi)$ We can find $c\cdot sin(\alpha+\frac{\pi}{2})>1 \Rightarrow cos\alpha>\frac{1}{c}$
Let x=a+t , t$\in(0,\frac{\pi}{2})$.
W=sin(x-$\alpha$)-c$\cdot$sinx=$sint-csin(a+t)<sint-\frac{sin(a+t)}{cosa}=\frac{sint\cdot cosa-sin(a+t)}{cosa}=\frac{-sina\cdot cost}{cosa}<0$
Therefore, two functions do not have any intersection point on ($a,a+\frac{\pi}{2}$).
(ii) $f(\alpha+\frac{\pi}{2})>g(\alpha+\frac{\pi}{2})$.
According to the monotonicity， there exists only one intersection point on $(\frac{\pi}{2},a+\frac{\pi}{2}).$
We can also find $c<\frac{1}{cosa}$
(1) When x $\in(a+\frac{\pi}{2},\pi)$.
Since $c<\frac{1}{cosa}$ , $W>sin(x-a)-\frac{sinx}{cosa}$
Let x=a+t, $t\in (\frac{\pi}{2},\pi-a)$
$W>sint-\frac{sin(a+t)}{cosa}=\frac{-sina\cdot cost}{cosa}$
Since $cosa>0 , cost<0$ , $W>\frac{-sina\cdot cost}{cosa}>0$ holds good on $\in(a+\frac{\pi}{2},\pi)$.
Therefore, two functions do not have any intersection point on $\in(a+\frac{\pi}{2},\pi)$.
(2) When x $\in(a,\frac{\pi}{2})$.
According to $f(\frac{\pi}{2})<g(\frac{\pi}{2})$ , it is clear that cosa<c.
$W=(cos\alpha-c)sinx-cosx\cdot sin\alpha<-cosx\cdot sin\alpha<0$
Therefore, two functions do not have any intersection point on $\in(a,\frac{\pi}{2})$.

The solution is not brief or beautiful at all. Waiting for better solutions.
