Let θ, φ ∈ [0, 2π] be such that 2cosθ(1 - sinφ) = sin2θ(tan(θ/2) + cot(θ/2))cosφ, tan(2π - θ) > 0 and -1 < sinθ < -√3/2. Then φ cannot satisfy which of the following intervals?

2cosθ(1 - sinφ) = sin2θ(tan(θ/2) + cot(θ/2))cosφ
⇒ 2cosθ(1 - sinφ) = sin2θ(2cscθsecθ)cosφ
⇒ 2cosθ(1 - sinφ) = 2sinθcosθ(2/sinθcosθ)cosφ
⇒ 1 - sinφ = 2cosφ
⇒ 1 = 2cosφ + sinφ
⇒ 1 = √5(2/√5 cosφ + 1/√5 sinφ)
Let cosα = 2/√5 and sinα = 1/√5, then tanα = 1/2.
⇒ 1 = √5 cos(φ - α)
⇒ cos(φ - α) = 1/√5
Since tan(2π - θ) > 0, this means θ is in the third quadrant.
Also, -1 < sinθ < -√3/2 which means θ ∈ (7π/6, 11π/6).
We know that α = arctan(1/2) ≈ 0.4636 radians which is approximately 26.57°.
Therefore, φ - α = ± arccos(1/√5) + 2nπ
φ ≈ α ± arccos(1/√5) + 2nπ
φ ≈ 0.4636 ± 1.107 + 2nπ
φ ≈ 1.5706 or -0.6434 + 2nπ
Since φ ∈ [0, 2π], φ can take values 1.5706, 4.7122, and 2.4976. We are given that -1 < sinθ < -√3/2, so we must choose the solution to 1 = 2cosφ + sinφ which will satisfy the given condition.
From 1 = 2cosφ + sinφ, we have
1 = √5(2/√5 cosφ + 1/√5 sinφ)
1/√5 = (2/√5) cosφ + (1/√5) sinφ
Let α be an angle such that cosα = 2/√5, sinα = 1/√5
Then we have cos(φ - α) = 1/√5
φ - α = ± arccos(1/√5) + 2kπ for some integer k.
Since 0 ≤ φ ≤ 2π, we must have 0 ≤ φ ≤ 2π. Then only 3π/2 < φ < 2π is not possible.