If $\sin ^2 \theta-3 \sin \theta+2=0$, then find the value of $\theta(0° \leq \theta \leq 90°)$.

45° 0° 60° 90°

90°

let sin\(\theta \) = x $\sin ^2 \theta-3 \sin \theta+2=0$ x² -3x + 2 = 0  Solving, (x - 2)(x - 1) = 0 \ x = 1. 2 since sin\(\theta \) \(\leq \) 1 sin\(\theta \) = 1 or \(\theta \) = 90°