For what value of θ, 1 lies between the roots of the quadratic equation 3x2 – 3sinθ x – 2cos2θ = 0?

For what value of θ, 1 lies between the roots of the quadratic equation 3x2 – 3sinθ x – 2cos2θ = 0? Correct Answer 2nπ + π/6 < θ < 2nπ + 5π/6

Let, f(x) = 3x2 – 3sinθ x – 2cos2θ The coefficient of x2 > 0 f(1) < 0 So, 3 – 3sinθ – 2cos2θ < 0 => 2sin2θ – 3sinθ + 1 < 0 => (2sinθ – 1)(sinθ – 1) < 0 => ½ 2nπ + π/6 < θ < 2nπ + 5π/6
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