Answer
$$\left\{ {{{57.7}^ \circ },{{159.2}^ \circ }} \right\}$$
Work Step by Step
$$\eqalign{
& \cot \theta + 2\csc \theta = 3 \cr
& {\text{Using identities}} \cr
& \frac{{\cos \theta }}{{\sin \theta }} + \frac{2}{{\sin \theta }} = 3 \cr
& \cos \theta + 2 = 3\sin \theta \cr
& {\text{Square each side}} \cr
& {\text{co}}{{\text{s}}^2}\theta + 4\cos \theta + 4 = 9{\sin ^2}\theta \cr
& {\text{co}}{{\text{s}}^2}\theta - 9{\sin ^2}\theta + 2\cos \theta + 4 = 0 \cr
& {\text{co}}{{\text{s}}^2}\theta - 9\left( {1 - {{\cos }^2}\theta } \right) + 4\cos \theta + 4 = 0 \cr
& {\text{co}}{{\text{s}}^2}\theta - 9 + 9{\cos ^2}\theta + 4\cos \theta + 4 = 0 \cr
& {\text{10co}}{{\text{s}}^2}\theta + 4\cos \theta - 5 = 0 \cr
& {\text{Let }}x = \cos \theta \cr
& {\text{10}}{x^2} + 4x - 5 = 0 \cr
& {\text{Solve by using the quadratic formula}} \cr
& x = \frac{{ - 4 \pm \sqrt {{{\left( {10} \right)}^2} - 4\left( {10} \right)\left( { - 5} \right)} }}{{2\left( {10} \right)}} \cr
& x = \frac{{ - 2 \pm 3\sqrt 6 }}{{10}} \cr
& {\text{,then}} \cr
& \cos \theta = \frac{{ - 2 - 3\sqrt 6 }}{{10}}{\text{ or }}\cos \theta = \frac{{ - 2 + 3\sqrt 6 }}{{10}} \cr
& {\text{Solve each equation}} \cr
& \theta = {\cos ^{ - 1}}\left( {\frac{{ - 2 - 3\sqrt 6 }}{{10}}} \right)\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\theta = {\cos ^{ - 1}}\left( {\frac{{ - 2 + 3\sqrt 6 }}{{10}}} \right)\, \cr
& \theta \approx {159.2^ \circ }\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\,\,\,\,\,\theta \approx {57.7^ \circ } \cr
& {\text{The solution set is}} \cr
& \theta \approx \left\{ {{{57.7}^ \circ },{{159.2}^ \circ }} \right\} \cr} $$
