JEE Advanced Mathematics Question (2023) — Solution

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 × 7   sin θ + 281 × 7 × 3   cos θ = 0 ⇒ 42 sin θ + 21 cos θ = 0 ⇒ tan θ = - 1 2 ⇒ cos 2 θ = 4 5   &   sin θ   cos θ = - 2 5 Now putting the value in  z we get, z = 281 × 7 × 7 - 281 × 6 × 3 × - 2 5 49 + 9 × 4 5 ⇒ z = 281 49 + 36 5 49 + 36 5 = 281 Hence, the value of  n  is  281