If sin ^2 θ-3 sin θ+2=0, then find the value of θ(0° ≤ θ ≤ 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°