Chapter 6 - Section 6.1 - Solving Trigonometric Equations - 6.1 Problem Set - Page 326: 37

$$\eqalign{ & \left( {\text{a}} \right)120^\circ + 360^\circ k,{\text{ }}240^\circ + 360^\circ k \cr & \left( {\text{b}} \right)120^\circ ,{\text{ }}240^\circ \cr} $$

$$\eqalign{ & 2{\cos ^2}\theta + 11\cos \theta = - 5 \cr & {\text{Add 5 to both sides of the equation}} \cr & 2{\cos ^2}\theta + 11\cos \theta + 5 = 0 \cr & {\text{Factoring }} \cr & \left( {2\cos \theta + 1} \right)\left( {\cos \theta + 5} \right) = 0 \cr & {\text{Use zero - factor property set each factor to 0}} \cr & 2\cos \theta + 1 = 0{\text{ or }}\cos \theta + 5 = 0 \cr & \cos \theta = - \frac{1}{2},{\text{ or }}\underbrace {\cos \theta = - 5}_{{\text{No solution}}} \cr & \cos \theta = - \frac{1}{2} \cr & {\text{The reference angle is given by}} \cr & \theta ' = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) \cr & \theta ' = 60^\circ \cr & {\text{The sign of }}\cos \theta {\text{ is negative, }}\theta {\text{ must terminate in the}} \cr & {\text{quadrants II or III}}{\text{. }} \cr & \theta = 180^\circ - 60^\circ {\text{ or }}\theta = 180^\circ + 60^\circ \cr & \theta = 120^\circ {\text{ or }}\theta = 240^\circ \cr & \cr & \left( {\text{a}} \right){\text{All the degree solutions are}} \cr & \theta = 120^\circ + 360^\circ k{\text{ or }}\theta = 240^\circ + 360^\circ k \cr & {\text{where }}k{\text{ is any integer}}{\text{.}} \cr & \cr & \left( {\text{b}} \right){\text{For 0}}^\circ \leqslant \theta \leqslant 360^\circ ,{\text{ the solution is in QII and QIII, then}} \cr & \theta = 120^\circ {\text{ or }}\theta = 240^\circ \cr} $$