2√3cos²θ = sinθ
find the value of θ
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Answer:
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Answer:
⇒ 2√3 cos² θ = sin θ
⇒ cos² θ / sinθ = 1/2√3
⇒ (1 - sin²θ) / sinθ = 1/2√3
Squaring both sides
⇒ (1 - sin²θ)² / sin²θ = (1/2√3)²
⇒ (1 - sin²θ)² / sin²θ = 1/12
⇒ 12 ( 1² + [sin²θ]² - 2sin²θ ) = sin²θ
⇒ 12 ( 1 + θ - 2sin²θ ) = sin²θ
⇒ 12 + 12θ - 24sin²θ = sin²θ
⇒ 12 + 12θ - 24sin²θ - sin²θ = 0
⇒ 12θ - 25sin²θ + 12 = 0
Taking sin²θ = y
⇒ 12y² - 25y + 12 = 0
⇒ 12y² - 16y - 9y + 12 = 0
⇒ 4y(3y-4) - 3(3y-4) = 0
⇒ (4y-3) (3y-4) = 0
Therefore, y = 3/4 Or
y = 4/3
So, sin²θ = 3/4 Or
sin²θ = 4/3
Taking sin²θ = 4/3
⇒ sin²θ = 4/3
⇒ sinθ = √4/√3
⇒ sinθ = 2/√3 Ignoring this as this will
not give the value of θ
Therefore, taking sin²θ = 3/4
⇒ sin²θ = 3/4
⇒ sinθ = √3/√4
⇒ sinθ = √3/2
And we know that sin60° = √3/2
Therefore, the value of θ = 60°
Hope it helps you .....
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