## Trigonometry 7th Edition

$\theta=60^{\circ},126^{\circ},233^{\circ},300^{\circ}$
$10(\cos\theta)^{2}+\cos\theta-3=0$ or, $10(\cos\theta)^{2}+6\cos\theta-5\cos\theta-3=0$ or, $(5\cos\theta+3)(2\cos\theta-1)=0$ So, $\cos\theta=-\frac{3}{5}\,or\,\frac{1}{2}$ Since, $0^{\circ}\leq\,\theta\lt360^{\circ}$ When $\cos\theta=-\frac{3}{5}$, $\theta=126^{\circ},233^{\circ}$ When $\cos\theta=\frac{1}{2}$, $\theta=60^{\circ},300^{\circ}$ So, the final set of solutions is $60^{\circ},126^{\circ},233^{\circ},300^{\circ}$