Answer
$$\theta = \frac{\pi }{4} + 2n\pi {\text{ and }}\theta = \frac{{3\pi }}{4} + 2n\pi ,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots $$
Work Step by Step
$$\eqalign{
& \sqrt 2 \sin x - 1 = 0 \cr
& {\text{Add 1 to both sides of the equation}} \cr
& \sqrt 2 \sin x = 1 \cr
& {\text{Divide both sides by }}\sqrt 2 \cr
& \sin x = \frac{1}{{\sqrt 2 }} \cr
& {\text{The equation }}\sin x = \frac{1}{{\sqrt 2 }}{\text{ has solutions }}\theta = \frac{\pi }{4}{\text{ and }}\theta = \frac{{3\pi }}{4}{\text{ in the }} \cr
& {\text{interval }}\left[ {0,2\pi } \right).{\text{ Moreover, because }}\cos \theta {\text{ has a period of 2}}\pi ,{\text{ there }} \cr
& {\text{are infinitely many other solutions, which can be written as}} \cr
& \theta = \frac{\pi }{4} + 2n\pi {\text{ and }}\theta = \frac{{3\pi }}{4} + 2n\pi \cr
& {\text{Where }}n{\text{ is an integer}}{\text{.}} \cr
& {\text{The general solution is:}} \cr
& \theta = \frac{\pi }{4} + 2n\pi {\text{ and }}\theta = \frac{{3\pi }}{4} + 2n\pi ,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots \cr} $$
