If 0 \leq x \leq \frac{\pi}{2}, then what is the number of values of x satisfying the equation \tan~x+\sec~x=2~\cos~x?
- A. 0
- B. 1 ✓
- C. 2
- D. 3
Correct Answer: B. 1
Explanation
Rewrite as \frac{\sin x + 1}{\cos x} = 2\cos x, which gives 1 + \sin x = 2\cos^2 x. Substituting \cos^2 x = 1 - \sin^2 x results in 2\sin^2 x + \sin x - 1 = 0. Factoring gives (2\sin x - 1)(\sin x + 1) = 0. Since 0 \leq x \leq \frac{\pi}{2}, \sin x must be non-negative. Hence, \sin x = \frac{1}{2}, yielding exactly one solution, x = \frac{\pi}{6}.