The vectors $\vec a (x) = \cos x\hat i + \sin x\hat j$ and $\vec b (x) = x\hat i + \sin x\hat j$ are collinear for

unique value of x, $\frac{π}{6}<x<\frac{π}{3}$

$\vec a (x)$ and $\vec b (x)$ will be collinear if and only if

$\cos x = x$

Let $f(x) = x - \cos x$. Then,

$f'(x)=1+ \sin x > 0$ for all x

$⇒ f(x)$ is an increasing function

$⇒f(x)=0$ for a unique value of x

Clearly, $f (x) > 0$ for $x>\frac{π}{3}$ and $f (x) <0$ for $x <\frac{π}{6}$

Thus, $f (x) = 0$ for a unique value of $x ∈ (π/6, π/3)$.

Hence, $\cos x = x$ for a unique value of $x ∈ (π/6, π/3)$.