The vectors $\overrightarrow{a}\left(x\right)=\cos x\widehat{i}+\sin x\widehat{j}$ and $\overrightarrow{b}\left(x\right)=x\widehat{i}+\sin x\widehat{j}$ are collinear for

$\overrightarrow{a}\left(x\right)$ and $\overrightarrow{b}\left(x\right)$ will be collinear if and only if $\cos x=x$

Let $f\left(x\right)=x-\cos x.$ Then,

$f^{\mathrm{\prime }}\left(x\right)=1+\sin x>0$ for all $x$

$\Rightarrow f\left(x\right)$ is an increasing function

$\Rightarrow f\left(x\right)=0$ for a unique value of $x$

Clearly., $f\left(x\right)>0\text{ for }x>\frac{\pi }{3}\text{ and }f\left(x\right)<0\text{ for }x<\frac{\pi }{6}$

$\text{ Thus, }f\left(x\right)=0\text{ for a unique value of }x\in (\pi /6,\pi /3)\text{. }$

$\text{ Hence, }\cos x=x\text{ for a unique value of }x\in (\pi /6,\pi /3)\text{. }$