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Question
If 2sin2θ – cos2θ = 2, then find the value of θ.
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Solution
Given,
2sin2θ – cos2θ = 2
⇒ 2sin2θ – (1 – sin2θ) = 2 ...[∵ sin2θ + cos2θ = 1]
⇒ 2sin2θ + sin2θ – 1 = 2
⇒ 3sin2θ = 3
⇒ sin2θ = 1
⇒ sinθ = 1 = sin 90° ...[∵ sin 90° = 1]
∴ θ = 90°
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