What is the smallest positive \( x \) satisfying \( \log_{\sin x} \cos x + \log_{\cos x} \sin x = 2 \) ?
- A. \frac{\pi}{2}
- B. \frac{\pi}{3}
- C. \frac{\pi}{4} ✓
- D. \frac{\pi}{6}
Correct Answer: C. \frac{\pi}{4}
Explanation
Let y = \log_{\sin x} \cos x. The equation becomes y + 1/y = 2, which gives y = 1. So \log_{\sin x} \cos x = 1, leading to \sin x = \cos x. The smallest positive x satisfying this is \pi/4.
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