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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