The number of elements in the set S = θ ∈ [ 0 , 2 π ] : 3 cos 4 θ - 5 cos 2 θ - 2 sin 6 θ + 2 = 0 is

Step-by-step solution

Given, 3 cos 4 θ - 5 cos 2 θ - 2 sin 6 θ + 2 = 0 ⇒ cos 2 θ 3 cos 2 θ - 5 - 2 sin 6 θ + 2 = 0 ⇒ 1 - sin 2 θ 3 - 3 sin 2 θ - 5 - 2 sin 6 θ + 2 = 0 ⇒ sin 2 θ - 1 3 sin 2 θ + 2 - 2 sin 6 θ + 2 = 0 Let sin 2 θ = t ⇒ ( t - 1 ) ( 3 t + 2 ) - 2 t 3 + 2 = 0 ⇒ ( t - 1 ) ( 3 t + 2 ) - 2 t 3 - 1 = 0 We know that a 3 - b 3 = a - b a 2 + a b + b 2 ⇒ ( t - 1 ) ( 3 t + 2 ) - 2 t - 1 t 2 + t + 1 = 0 ⇒ ( t - 1 ) 3 t + 2 - 2 t 2