Answer
$$\theta = 0,\,\,\,\,\,\theta = \frac{{2\pi }}{3} + 2n\pi {\text{ }}\,\,\,\,{\text{and }}\,\,\,\,\theta = \frac{{4\pi }}{3} + 2n\pi $$
Work Step by Step
$$\eqalign{
& 2\theta \cos \theta + \theta = 0 \cr
& {\text{Factoring}} \cr
& \theta \left( {2\cos \theta + 1} \right) = 0 \cr
& {\text{Use the zero - factor property}} \cr
& \theta = 0{\text{ or }}2\cos \theta + 1 = 0 \cr
& {\text{Therefore,}} \cr
& 2\cos \theta + 1 = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\cos \theta = - \frac{1}{2} \cr
& {\text{The equation }}\cos \theta = - \frac{1}{2}{\text{ has solutions }}\theta = \frac{{2\pi }}{3}{\text{ and }}\theta = \frac{{4\pi }}{3}{\text{ in the }} \cr
& {\text{interval }}\left[ {0,2\pi } \right).{\text{ Moreover, because }}\cos \theta {\text{ has a period of 2}}\pi ,{\text{ there }} \cr
& {\text{are infinitely many other solutions, which can be written as}} \cr
& \theta = \frac{{2\pi }}{3} + 2n\pi {\text{ and }}\theta = \frac{{4\pi }}{3} + 2n\pi \cr
& {\text{Where }}n{\text{ is an integer}}{\text{.}} \cr
& {\text{The general solution is:}} \cr
& \theta = 0,\,\,\,\,\,\theta = \frac{{2\pi }}{3} + 2n\pi {\text{ }}\,\,\,\,{\text{and }}\,\,\,\,\theta = \frac{{4\pi }}{3} + 2n\pi \cr} $$
