Find exact values for all the solutions to the equation \( \sin \theta = \frac{-\sqrt{2}}{2} \)

5th Ed: #29

Since the value is negative, we know that \( \theta \) is in the III or IV quadrant.
So limiting ourselves to \( 0 \leq \theta \leq 2 \pi \), we would have \( \theta = \frac{5\pi}{4},\frac{7\pi}{4} \).
Now opening up \( \theta \) to all real numbers, we would have \( \theta = \frac{5\pi}{4}+2\pi k,\frac{7\pi}{4}+2\pi k \) where \( k \) is an integrer.