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The number of solution(s) of 2sin2θ−cos2θ=0 and 2cos2θ−3sinθ=0 for θ∈[0,2π] is

Answer» The number of solution(s) of 2sin2θcos2θ=0 and 2cos2θ3sinθ=0 for θ[0,2π] is


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