The eccentric angle of a point on the ellipse $x^2+3y^2=6$ at a distance 2 unit from origin is

$$\begin{gathered} x^2+3y^2=6 \\ \frac{x^2}{6}+\frac{y^2}{2}=1 \\ {a}^2=6,b^2=2 \end{gathered}$$

$OP=\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }$

$2=\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }$

$6{\cos }^2\theta +2{\sin }^2\theta =4$

$6{\cos }^2\theta +2(1-{\cos }^2\theta )=4$

$6{\cos }^2\theta +2-2{\cos }^2\theta =4$

$$\begin{gathered} 4{\cos }^2\theta =2 \\ {\cos }^2\theta =\frac{1}{2} \end{gathered}$$

$\cos \theta =\pm \frac{1}{\sqrt{2}}$

$\theta =\frac{n\pi }{2}+\frac{\pi }{4}=(2n+1)\frac{\pi }{4}$

$\theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$