Let $f\left(x\right)=a\cos x+b\sin x.$ If there exist $x_1,x_2\text{ such that }f\left(x_1\right)=f\left(x_2\right)=0$ and $x_1-x_2\ne n\pi ,$ $n\in \mathrm{\mathbb{Z}}$ then the number of common soluton(s) of $y=f\left(x\right)$ and $y=\prod _{k=1}^5 (x-k-a)(x+k-b)$ is ______

$\text{ claim: }a=b=0$

$\text{ If possible let me of }a,b\ne 0$

$y=f\left(x\right)=a\cos x+b\sin x$

$=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\right)$

$=\sqrt{a^2+b^2}\sin (x+\varphi )\text{ where }\cos \varphi =\frac{b}{\sqrt{a^2+b^2}}\text{ and }$

$\sin \varphi =\frac{a}{\sqrt{a^2+b^2}}$

$\text{ Now }f\left(x_1\right)=0\Rightarrow x_1+\varphi =n_1\pi ,n_1\in \mathrm{\mathbb{Z}}$

$f\left(x_2\right)=0\Rightarrow x_2+\varphi =n_2\pi ,n_2\in \mathrm{\mathbb{Z}}$

$\therefore x_1-x_2=n\pi$

But it is given that $x_1-x_2\ne n\pi ,n\in \mathrm{\mathbb{Z}}$

$\text{ Hence }a=b=0$

$\therefore y=f\left(x\right)=0\text{ and }y=\prod _{k=1}^5 (x-k-a)(x+k-b)$

$=\prod _{k=1}^5 (x-k)(x+k)$

Hence the number of common solutions = 10.