The quadratic equation 1 - sin ⁡ θ x 2 + 2 1 - sin ⁡ θ x - 3 sin ⁡ θ = 0 has both roots complex for all θ lying in the interval

Step-by-step solution

D = 4 1 - sin ⁡ θ 2 + 4 1 - sin ⁡ θ ⋅ 3 sin ⁡ θ < 0 ⇒ 4 1 - sin ⁡ θ 1 - sin ⁡ θ + 3 sin ⁡ θ < 0 ⇒ 4 1 - sin ⁡ θ 1 + 2 sin ⁡ θ < 0 ⇒ 8 sin ⁡ θ - 1 sin ⁡ θ + 1 2 > 0 ⇒ sin ⁡ θ < - 1 2 o r sin ⁡ θ > 1 sin ⁡ θ < - 1 2 ⇒ θ ∈ 7 π 6 , 11 π 6