Answer
$\{19.5^{o}+360^{o}k, 160.5^{o}+360^{o}k,$ $210^{o}+360^{o}k, 330^{o}+360^{o}k, \ \ $ (k any integer)$\}$
Work Step by Step
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,\ \ $(k any integer)$\}$
