??
Hello, astrostar
You're almost there . . .
$\displaystyle 1 -2\sin^2\!\theta + \cos^2\!\theta \:= \:0\;\text{ for }0^o \leq \theta < 360^o$
My answer so far:
. . $\displaystyle 1 + \sin^2\!\theta + \cos^2\!\theta \:= \:3\sin^2\!\theta$
. . $\displaystyle 2 \:= \:3\sin^2\!\theta$ . . . . Good!
We have: .$\displaystyle 3\sin^2\!\theta \:=\: 2\quad\Rightarrow\quad \sin^2\!\theta \:=\:\frac{2}{3}\quad\Rightarrow\quad \sin\theta \:=\:\pm\sqrt{\frac{2}{3}}$
Therefore: .$\displaystyle \theta \;\;=\;\;\sin^{-1}\left(\pm\sqrt{\frac{2}{3}}\right) \;\;\approx\;\; \left\{54.7^o,\:125.3^o,\:234.7^o,\:305.3^o\right\ }$
You were on the right track; just continue with what you were doing!
$\displaystyle 2=3sin^2(E)$
$\displaystyle sin^2(E)=\frac{2}{3}$
$\displaystyle sin(E)=\pm \sqrt{\frac{2}{3}}$
$\displaystyle E = sinh(\pm \frac{2}{3})$
You'll have to use your calculator for the last part!