The number of values of θ ∈ − 3 π 2 , 4 π 3 which satisfies the system of equations 2 sin 2 θ + sin 2 2 θ = 2 and sin 2 θ + cos 2 θ = tan θ is

Step-by-step solution

First equation is 4 sin 2 θ cos 2 θ = 2 cos 2 θ ⇒ cos 2 ⁡ θ = 0 or sin 2 θ = 1 2 = 1 2 2 = sin 2 π 4 ⇒ θ = 2 n + 1 π 2 , n ∈ I or θ = n π ± π 4 , n ∈ I Second equation is not satisfied by θ = 2 n + 1 π 2 , n ∈ I but satisfied by θ = n π ± π 4 , n ∈ I So θ = nπ ± π 4 , n ∈ I ∴ In [ − 3 π 2 , 4 π 3 ] , the values of θ are − 5 π 4 , − 3 π 4 , − π 4 , π 4 , 3 π 4 , 5 π 4