## Precalculus (6th Edition)

$\{19.5^{o}+360^{o}k, 160.5^{o}+360^{o}k,$ $210^{o}+360^{o}k, 330^{o}+360^{o}k, \ \$ (k any integer)$\}$
substitute t=$\sin\theta,$ $6t^{2}+t-1=0\qquad$... quadratic formula... $t=\displaystyle \frac{-1\pm\sqrt{1+24}}{2\cdot 6}=\frac{-1\pm 5}{12}$ $t_{1}=\displaystyle \frac{-1-5}{12}=-\frac{1}{2}$ $t_{2}=\displaystyle \frac{-1+5}{12}=\frac{1}{3}$ 1. $\displaystyle \sin\theta=-\frac{1}{2}$, sine is negative in quadrants III and IV, From the unit circle: $\theta$ can be $210^{o}$ or $330^{o}$ The period for sine is $360^{o}$, so the solutions here are $\{210^{o}+360^{o}k, 330^{o}+360^{o}k,$(k any integer)$\}$ 2. $\displaystyle \sin\theta=\frac{1}{3}$, sine is positive in quadrants I and II, Calculator: $\theta$ can be $19.5^{o}$ or $180^{o}-19.5^{o}=160.5^{o}$ The period for sine is $360^{o}$, so the solutions here are $\{19.5^{o}+360^{o}k, 160.5^{o}+360^{o}k,$(k any integer)$\}$ Solution set: $\{19.5^{o}+360^{o}k, 160.5^{o}+360^{o}k,$ $210^{o}+360^{o}k, 330^{o}+360^{o}k,\ \$(k any integer)$\}$