Let A = \ (1967 + 1686isinθ)/(7 - 3icosθ): θ ∈ R\ . If A contains...

Let $z = \dfrac{1967 + 1686isin\theta}{7 - 3icos\theta}$ is a positive integer.

$⇒ z = \dfrac{\left(1967 + 1686isin\theta\right)\left(7 + 3icos\theta\right)}{\left(7 - 3icos\theta\right)\left(7 + 3icos\theta\right)}$

$⇒ z = \dfrac{1967 \times 7 - 1686 \times 3sin\theta cos\theta + i\left(1686 \times 7sin\theta + 1967 \times 3cos\theta\right)}{49 + 9\cos^{2}\theta}$

Now taking imaginary part as zero we get,

$1686 \times 7sin\theta + 1967 \times 3cos\theta = 0$

$⇒ \left(281 \times 6\right) \times 7sin\theta + \left(281 \times 7\right) \times 3cos\theta = 0$

$⇒ 42sin\theta + 21cos\theta = 0$

$⇒ tan\theta = - \dfrac{1}{2}$

$⇒ \cos^{2}\theta = \dfrac{4}{5} \text{ and } sin\theta cos\theta = - \dfrac{2}{5}$

Now putting the value in $z$we get,

$z = \dfrac{\left(281 \times 7 \times 7\right) - \left(281 \times 6\right) \times 3 \times \left(- \dfrac{2}{5}\right)}{49 + 9 \times \dfrac{4}{5}}$

$⇒ z = \dfrac{281\left(49 + \dfrac{36}{5}\right)}{\left(49 + \dfrac{36}{5}\right)} = 281$

Hence, the value of $n$ is $281$