If θ 1 and θ 2 be respectively the smallest and the largest values of θ in 0 , 2 π - π which satisfy the equation, 2 cot 2 ⁡ θ - 5 sin ⁡ θ + 4 = 0 , then ∫ θ 1 θ 2 cos 2 ⁡ 3 θ d θ is equal to:

Step-by-step solution

2 cot 2 ⁡ θ - 5 sin ⁡ θ + 4 = 0 2 cos 2 ⁡ θ sin 2 ⁡ θ - 5 sin ⁡ θ + 4 = 0 2 cos 2 ⁡ θ - 5 sin ⁡ θ + 4 sin 2 ⁡ θ = 0 ,   sin ⁡ θ ≠ 0 2 sin 2 ⁡ θ - 5 sin ⁡ θ + 2 = 0 2 sin ⁡ θ - 1 sin ⁡ θ - 2 = 0 sin ⁡ θ = 1 2 θ = π 6 , 5 π 6 ; θ ∈ 0 , 2 π - π ∴ ∫ π / 6 5 π / 6 cos 2 ⁡ 3 θ d θ = &#8