\( 2 \cos ^ { 2 } ( 45 ^ { \circ } - \theta ) = 1 + \sin 2 \theta \)

To solve the equation \( 2 \cos^2 (45^\circ - \theta) = 1 + \sin 2\theta \), we can start by using trigonometric identities. First, remember that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Now, substituting it in, we get: \[ 2 \cos^2 (45^\circ - \theta) = 1 + 2 \sin \theta \cos \theta. \] Next, we can simplify \( \cos^2 (45^\circ - \theta) \) using the cosine subtraction formula: \[ \cos (45^\circ - \theta) = \frac{\sqrt{2}}{2} (\cos \theta + \sin \theta). \] Squaring it gives us: \[ \cos^2 (45^\circ - \theta) = \frac{1}{2} (\cos^2 \theta + 2 \sin \theta \cos \theta + \sin^2 \theta) = \frac{1}{2} (1 + 2 \sin \theta \cos \theta). \] Substituting this back into the original equation: \[ 2 \left( \frac{1}{2} (1 + 2 \sin \theta \cos \theta) \right) = 1 + 2 \sin \theta \cos \theta. \] This simplifies to: \[ 1 + 2 \sin \theta \cos \theta = 1 + 2 \sin \theta \cos \theta. \] Since both sides are equivalent, the equation holds true for all values of \( \theta \). Therefore, the solution is: All real numbers \( \theta \).