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$If 4 x^{2} + 2 \cos^{2} θ = 4 x - \sin^{2} θ then the maximum number of ordered pairs (x, θ) in [0, 2 π] is$

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Correct option is C

4x2−4x+1+cos2⁡θ=0⇒ (2x−1)2+cos2⁡θ=0⇒ 2x−1=0 and cos⁡θ=0⇒ x=12 and θ=π2,3π2⇒ 12,π2;12,3π2Thus, 2 ordered pairs are possible.

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If 4x2+2cos2⁡θ=4x−sin2⁡θ then the maximum number of ordered pairs (x,θ) in [0,2π] is

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$If 4 x^{2} + 2 \cos^{2} θ = 4 x - \sin^{2} θ then the maximum number of ordered pairs (x, θ) in [0, 2 π] is$