What is the number of solutions of (sinθ – cosθ)² = 2 where –π < θ < π?

(sinθ – cosθ)² = sin²θ – 2sinθcosθ + cos²θ = 1 – sin2θ = 2, so sin2θ = –1. Then 2θ = –π/2 + 2kπ or 2θ = 3π/2 + 2kπ (same family). So θ = –π/4 + kπ. In (–π, π): θ = –π/4 and θ = 3π/4. Two solutions.