Question

For \( a \in \mathbb{R} \setminus \{0\} \), if \( a \cos x + a \sin x + a = 2K + 1 \) has a solution, then \( K \) lies in the interval:

• A. \( \frac{a - 1 - \sqrt{2a}}{2}, \frac{a - 1 + \sqrt{2a}}{2} \)
• B. \( \frac{a + 1 - \sqrt{2a}}{2}, \frac{a + 1 + \sqrt{2a}}{2} \)
• C. \( \frac{a - 1 - \sqrt{2a + 2a + 1}}{2}, \frac{a - 1 + \sqrt{2a + 2a + 1}}{2} \)
• D. \( \frac{\sqrt{2a^2 + 2a + 1} + 1}{2}, \frac{\sqrt{2a^2 + 2a + 1} - 1}{2} \)

Hint:

For solving trigonometric equations involving sums of sines and cosines, use the sum-to-product identities to simplify the equation and find the required range.

Answer 1

The Correct Option is A

We are given that: \[ a \cos x + a \sin x + a = 2K + 1. \] This equation can be rewritten as: \[ a (\cos x + \sin x) = 2K + 1 - a. \] We know that: \[ \cos x + \sin x = \sqrt{2} \sin \left( x + \frac{\pi}{4} \right). \] Thus, the equation becomes: \[ a \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) = 2K + 1 - a. \] Since \( \sin \left( x + \frac{\pi}{4} \right) \) can take values between -1 and 1, we can analyze the equation: \[ - a \sqrt{2} \leq 2K + 1 - a \leq a \sqrt{2}. \] Solving for \( K \), we get: \[ K = \frac{a - 1 - \sqrt{2a}}{2} \quad {to} \quad \frac{a - 1 + \sqrt{2a}}{2}. \] Thus, the value of \( K \) lies in the interval: \[ \boxed{ \frac{a - 1 - \sqrt{2a}}{2}, \frac{a - 1 + \sqrt{2a}}{2} }. \]