Answer
$$\eqalign{
& \left( {\text{a}} \right)30^\circ + 360^\circ k,{\text{ 15}}0^\circ + 360^\circ k \cr
& \left( {\text{b}} \right)30^\circ ,{\text{ 15}}0^\circ \cr} $$
Work Step by Step
$$\eqalign{
& 2{\sin ^2}\theta - 7\sin \theta = - 3 \cr
& {\text{Add 3 to both sides of the equation}} \cr
& 2{\sin ^2}\theta - 7\sin \theta + 3 = 0 \cr
& {\text{Factoring }} \cr
& \left( {\sin \theta - 3} \right)\left( {2\sin \theta - 1} \right) = 0 \cr
& {\text{Use zero - factor property set each factor to 0}} \cr
& \sin \theta - 3 = 0{\text{ or }}2\sin \theta - 1 = 0 \cr
& \underbrace {\sin \theta = 3}_{{\text{No solution}}},{\text{ or }}2\sin \theta = 1 \cr
& \sin \theta = \frac{1}{2} \cr
& {\text{The reference angle is given by}} \cr
& \theta ' = {\sin ^{ - 1}}\left( {\frac{1}{2}} \right) \cr
& \theta ' = 30^\circ \cr
& {\text{The sign of }}\sin \theta {\text{ is positive, }}\theta {\text{ must be terminate in the}} \cr
& {\text{quadrants I or II}}{\text{. }} \cr
& \theta = 30^\circ + 0^\circ {\text{ or }}\theta = 180^\circ - 30^\circ \cr
& \theta = 30^\circ {\text{ or }}\theta = 150^\circ \cr
& \cr
& \left( {\text{a}} \right){\text{All the degree solutions are}} \cr
& \theta = 30^\circ + 360^\circ k{\text{ or }}\theta = 150^\circ + 360^\circ k \cr
& {\text{where }}k{\text{ is any integer}}{\text{.}} \cr
& \cr
& \left( {\text{b}} \right){\text{For 0}}^\circ \leqslant \theta \leqslant 360^\circ ,{\text{ the solution is in QI and QII, then}} \cr
& \theta = 30^\circ {\text{ or }}\theta = 150^\circ \cr} $$
