Let A = {\(\frac{1967+1686 i\,sin\theta}{7-3i\,cos\theta}:\theta\in R\) }. If A contains exactly one positive integer n, then the value of n is ______
Correct Answer: 281
Solution and Explanation
The given expression is:
\(\frac{281.(7) + 218.(6i \, \sin(x))}{7 - 3i \, (\cos(x))}\)
We simplify this as:
\(\Rightarrow 281\left(\frac{7 + 6i \, \sin(x)}{7 - 3i \, \cos(x)}\right)\)
Next, we multiply both the numerator and denominator by \(7 + 3i \, \sin(x)\):
\(\Rightarrow 281\left(\frac{49 + 21i (\cos(x) + 2 \sin(x)) + 18 \sin(x) \cos(x)}{49 + 9 \cos^2(x)}\right)\)
This simplifies to:
\(\Rightarrow \left(\frac{49 + 18 \sin(x) \cos(x)}{49 + 9 \cos^2(x)} + i \frac{21 (\cos(x) + 2 \sin(x))}{49 + 9 \cos^2(x)}\right)\)
Hence, we can conclude that:
\(\frac{21 \cos(x) + 2 \sin(x)}{49 + 9 \cos^2(x)} = 0\)
Therefore, the value of \( n \) is 281.
Thus, the value of \( n \) is 281.
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Concepts Used:
Complex Number
A Complex Number is written in the form
a + ib
where,
- “a” is a real number
- “b” is an imaginary number
The Complex Number consists of a symbol “i” which satisfies the condition i^2 = −1. Complex Numbers are mentioned as the extension of one-dimensional number lines. In a complex plane, a Complex Number indicated as a + bi is usually represented in the form of the point (a, b). We have to pay attention that a Complex Number with absolutely no real part, such as – i, -5i, etc, is called purely imaginary. Also, a Complex Number with perfectly no imaginary part is known as a real number.