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

Question

• 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 ?

Answer 1

A necessary, but insufficient condition for a solution to exist is the fact that

$x^2+1=a^2+b^2$

Which can be obtained by squaring both equations and adding them together.

Most likely, you will need to analyse the equation with the aid of a graphical depiction, in order to determine sufficient conditions for solutions to exist.

Update: Dividing the initial equation by the derivative equation, we obtain:

$\displaystyle \frac{a\sin{x} + b\cos{x}}{a\cos{x}-b\sin{x}} = x$

$\displaystyle \frac{\tan{x}+\frac{b}{a}}{1-\frac{b}{a}\times\tan{x}} = x$

Which is equivalent to:

$\displaystyle \tan{ ( x+\tan^{-1}{\frac{b}{a})}} = x$

This definitely requires graphical aid, but it is significantly easier to handle.

Answer 2

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 $a$: $$a\pi+4b=b\pi-4a$$ $$a\pi+4a=b\pi-4b$$ $$b=a\dfrac{\pi+4}{\pi-4}$$

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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 $b$.