sin²θ– 3 sin θ + 2 = 0 will be true

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sin²θ– 3 sin θ + 2 = 0 will be true if

1. 0 ≤ θ < 90
2. 0 < θ < 90
3. θ = 0°
4. θ = 90°

Correct Option: D

sin² θ – 3 sin θ + 2 = 0
⇒ sin² θ – 2 sin θ – sin θ + 2 = 0
⇒ sin θ (sin θ – 2) –1 (sin θ – 2) = 0
⇒ (sin θ – 1) (sin θ – 2) = 0
⇒ sin θ = 1 = sin 90°
⇒ θ = 90° and sin θ ≠ 2