Consider the following lists: List- I List- II I x ∈ - 2 π 3 , 2 π 3 : cos x + sin x = 1 P has two elements II x ∈ - 5 π 18 , 5 π 18 : 3 tan 3 x = 1 Q has three elements III x ∈ - 6 π 5 , 6 π 5 : 2 cos 2 x = 3 R has four elements IV x ∈ - 7 π 4 , 7 π 4 : sin x - cos x = 1 S has five elements T has six elements The correct option is:

Step-by-step solution

Solving all questions one by one we get, (i) x ∈ - 2 π 3 , 2 π 3 , cos x + sin x = 1 cos x + sin x = 1 ⇒ sin π 4 + x = 1 2 ⇒ π 4 + x = n π + - 1 n π 4 ⇒ x = n π + - 1 n π 4 - π 4 So, x ∈ 0 , π 2 ∴     x has 2 elements. → P (ii) x ∈ - 5 π 18 , 5 π 18 : 3 tan 3 x = 1 Solving 3 tan 3 x = 1 ⇒ tan 3 x = 1 3 ⇒ 3 x = n π + π 6 ⇒ x = n π 3 + π 18 So, x ∈ π 18