#### Answer

$\{53.6^{o},126.4^{o},187.9^{o},352.1^{o}\}$

#### Work Step by Step

Substitute t=$\sin\theta$:
$ 9t^{2}-6t-1=0\qquad$ ... quadratic formula...
$t=\displaystyle \frac{6\pm\sqrt{36+36}}{18}=\frac{6\pm 6\sqrt{2}}{18}$
$t=\displaystyle \frac{6(1\pm\sqrt{2})}{6(3)}$
Back-substitute
$\displaystyle \sin\theta=\frac{1\pm\sqrt{2}}{3}$
In quadrant I, our calculator gives
$\displaystyle \sin^{-1}(\frac{1+\sqrt{2}}{3})\approx$53.5849460516$\approx 53.6^{o}$
In q.II, $180^{0}-53.6=126.4^{o}$
For the other (negative) value, the calculator gives
$\displaystyle \sin^{-1}(\frac{1-\sqrt{2}}{3})\approx$-7.93624951449$\approx-7.9^{o},$
which is not in the interval $[0,\ 360)$ ,
so we adjust with multiples of 180:
(sines are negative in quadrants III and IV)
In q. III, $180-(-7.9)=187.9^{o}$
In q.IV, $360^{o}-7.9=352.1^{o}$
Solution set = $\{53.6^{o},126.4^{o},187.9^{o},352.1^{o}\}$