#### Answer

$\{90^{o},210^{o},330^{o}\}$

#### Work Step by Step

Substituting t$=\sin\theta$ ,the equation becomes
$2t-1=\displaystyle \frac{1}{t}\qquad/\times t$
$2t^{2}-t=1$
$2t^{2}-t-1=0$
Using the quadratic formula,
$t=\displaystyle \frac{1\pm\sqrt{1+8}}{4}=\frac{1\pm 3}{4}$
$t=1$ or $ t=-\displaystyle \frac{1}{2}$
Back substitute, use the unit circle:
$\begin{array}{lll}
\sin\theta=1 & or & \sin\theta=-\frac{1}{2}\\
\theta=90^{o} & or & \theta=210^{o},330^{0}
\end{array}$
Solution set= $\{90^{o},210^{o},330^{o}\}$