Find the values of (0° ≤ θ ≤ 90° ) for which 2 sinθ cosθ = cosθ.
Solution :
2 sinθ cosθ = cosθ
| ⇒ 2 sinθ cosθ – cosθ = 0 ⇒ cosθ ( 2 sinθ – 1 ) = 0 Either cosθ = 0 ⇒ cosθ = cos90° ( ∴ cos90° = 0 ) ⇒ θ = 90° or 2 sinθ – 1 = 0 ⇒ 2 sinθ = 1 ⇒ sinθ = 1 / 2 ∴ sinθ = sin 30° [ ∵ sin30° = 1 / 2 ] ∴ θ = 30° ∴ The required values of θ are 30° and 90°. |
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