Let A = 1967 + 1686 i sin θ 7 - 3 i cos θ : θ ∈ R . If A contains exactly one positive integer n , then the value of n is

Step-by-step solution

Let z = 1967 + 1686 i   sin θ 7 - 3 i   cos θ is a positive integer. ⇒ z = 1967 + 1686 i   sin θ 7 + 3 i   cos θ 7 - 3 i   cos θ 7 + 3 i   cos θ ⇒ z = 1967 × 7 - 1686 × 3   sin θ   cos θ + i 1686 × 7   sin θ + 1967 × 3   cos θ 49 + 9   cos 2 θ Now taking imaginary part as zero we get, 1686 × 7   sin θ + 1967 × 3   cos θ = 0 ⇒ 281 × 6 &