When does the function $a\cdot \sin(x)+b\cdot \cos(x)-x$ have exactly one real root with multiplicity $1$? 
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*When does the function $$f(x)=a\cdot \sin(x)+b\cdot \cos(x)-x$$ have exactly one real root ($a$ and $b$ are real numbers) ?

*When does $f(x)$ have a multiple root , in other words for which real numbers $a$ and $b$ does the system $$a\cdot \sin(x)+b\cdot \cos(x)=x$$ $$a\cdot \cos(x)-b\cdot\sin(x)=1$$ have a real solution $x$ ?
A root with multiplicity $3$ occurs, if we have $a=1$ and $b=0$. The function $f(x)=\sin(x)-x$ and the first two derivates are $0$ for $x=0$.
If we choose $a=4$ and $b=-6$, we have $5$ real simple roots. The choice $a=1$ and $b=1$ leads to a unique simple root.
$f(x)$ has always a real root because of $$\lim_{x\rightarrow \pm \infty} f(x)=\mp \infty$$
Any ideas ?

Okay, so we have:
$$a\cdot\sin(x)+b\cdot\cos(x)=x$$
$$a\cdot\cos(x)-b\cdot\sin(x)=1$$
multiply bottom one by $\dfrac ab$
$$\dfrac{a^2}{b}\cdot\cos(x)-b\cdot\sin(x)=\dfrac ab$$
Add it to the top one:
$$(\dfrac{a^2+b^2}{b})\cos(x)=x+\dfrac{a}{b}$$
So:
$$\cos(x)=\dfrac{bx+a}{a^2+b^2}$$
Now multiply the bottom one by $-\dfrac{b}{a}$:
$$-b\cdot\cos(x)+\dfrac{b^2}{a}\cdot\sin(x)$$
Add it to the top one:
$$(\dfrac{a^2+b^2}{a})\cdot\sin(x)=x-\dfrac{b}{a}$$
So:
$$\sin(x)=\dfrac{ax-b}{a^2+b^2}$$
We now take:
$$\tan(x)=\dfrac{\sin(x)}{\cos(x)}=\dfrac{ax-b}{bx+a}$$
Now, we use $\tan(\dfrac{\pi}{4})=1$  to obtain:
$$\dfrac{a\pi+4b}{b\pi-4a}=1$$
Use that equation to get $b$ in terms of $a$:
$$a\pi+4b=b\pi-4a$$
$$a\pi+4a=b\pi-4b$$
$$b=a\dfrac{\pi+4}{\pi-4}$$

Now we have:
$$a(\sin(x)+\dfrac{\pi+4}{\pi-4}\cdot\cos(x))=x$$
$$a(\cos(x)-\dfrac{\pi+4}{\pi-4}\cdot\sin(x))=1$$
Divide the top one by the bottom one:
$$\dfrac{\sin(x)+\dfrac{\pi+4}{\pi-4}\cdot\cos(x)}{\cos(x)-\dfrac{\pi+4}{\pi-4}\sin(x)}=\dfrac{(\pi-4)\sin(x)+(\pi+4)\cos(x)}{(\pi-4)\cos(x)-(\pi+4)\sin(x)}=x$$
For every $x$ that is s solution to this equation, you can find $a$ and $b$.
