Given trigonometric relation is 

${\left(\frac{1}{3}\sin \theta +\frac{2}{3}\cos \theta \right)}^2=\frac{1}{3}{\sin }^2\theta +\frac{2}{3}{\cos }^2\theta$

$\Rightarrow \frac{1}{9}{\sin }^2\theta +\frac{4}{9}{\cos }^2\theta +\frac{4}{9}\sin \theta \cos \theta$

$=\frac{1}{3}{\sin }^2\theta +\frac{2}{3}{\cos }^2\theta$

$\Rightarrow \frac{2}{9}{\sin }^2\theta +\frac{2}{9}{\cos }^2\theta -\frac{4}{9}\sin \theta \cos \theta =0$

$\Rightarrow {\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta =0$

$\Rightarrow \sin 2\theta =1$

$\Rightarrow 2\theta =2n\pi +\frac{\pi }{2}, n\in I$

$\Rightarrow \theta =n\pi +\frac{\pi }{4}, n\in I$

$\therefore A=\left\{\theta \in R:\theta =n\pi +\frac{\pi }{4},n\in I\right\}$

$\therefore A\cap \left[0,\pi \right]=\left\{\frac{\pi }{4}\right\}$

$\therefore A\cap \left[0,\pi \right]$ has exactly one point.