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Question
If 2sin2θ = 3cosθ, where 0 ≤ θ ≤ 2π, then find the value of θ.
Solution
2sin2θ = 3cosθ
We know that,
sin2θ = 1 – cos2θ
Given that,
2sin2θ = 3 cosθ
2 – 2cos2θ = 3cosθ
2cos2θ + 3cosθ – 2 = 0
(cosθ + 2)(2cosθ – 1) = 0
Therefore,
cosθ = `1/2 = cos pi/3`
θ = `pi/3` or `2π – pi/3`
θ = `pi/3, (5pi)/3`
Therefore, 2(1 – cos2θ) = 3cosθ
⇒ 2 – 2cos2θ = 3cosθ
⇒ 2cos2θ + 3cosθ – 2 = 0
⇒ 2cos2θ + 4cosθ – cosθ – 2 = 0
⇒ 2cosθ(cosθ + 2) + 1(cosθ + 2) = 0
⇒ (2cosθ + 1)(cosθ + 2) = 0
Since, cosθ ∈ [–1, 1], for any value θ.
So, cosθ ≠ –2
Therefore,
2cosθ – 1 = 0
⇒ cosθ = `1/2`
= `pi/3` or `2π – pi/3`
θ = `π/3, (5pi)/3`
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