## Precalculus (6th Edition)

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