Question

If $ 0 \le x \le 3,\ 0 \le y \le 3 $, then the number of solutions $(x, y)$ for the equation: $$ \left( \sqrt{\sin^2 x - \sin x + \frac{1}{2}} \right)^{\sec^2 y} = 1 $$

• A. 5
• B. 2
• C. 6
• D. 1

Hint:

When evaluating equations with exponentials equaling 1, always check if base = 1 or exponent = 0, and solve accordingly.

Answer 1

The Correct Option is B

Let: \[ f(x, y) = \left( \sqrt{\sin^2 x - \sin x + \frac{1}{2}} \right)^{\sec^2 y} = 1 \] Case 1: Base is 1 → then \( \sqrt{\sin^2 x - \sin x + \frac{1}{2}} = 1 \) Square both sides: \[ \sin^2 x - \sin x + \frac{1}{2} = 1 \Rightarrow \sin^2 x - \sin x - \frac{1}{2} = 0 \Rightarrow \sin x = \frac{1 \pm \sqrt{3}}{2} \Rightarrow Acceptable values of \( x \in [0, 3] \Rightarrow \) only 1 solution Case 2: Exponent is 0 → \( \sec^2 y = 0 \) is not possible Case 3: Base is -1, exponent even (say 2) → not allowed as square root must be real So valid only when: \[ \sqrt{\sin^2 x - \sin x + \frac{1}{2}} = 1,\ \text{which gives two } x \text{ values} \Rightarrow 2 solutions overall \]