Given 1. \( C=\frac{\pi}{2}+\sqrt{\frac{2}{3+x}} \) for what value(s) of a will \( C \) be a real number?

To determine the values of \( x \) for which \( C \) is a real number, we need to analyze the expression: \[ C = \frac{\pi}{2} + \sqrt{\frac{2}{3+x}} \] The term \( \sqrt{\frac{2}{3+x}} \) must be a real number. For the square root to be defined and real, the expression inside the square root must be non-negative: \[ \frac{2}{3+x} \geq 0 \] This inequality holds true when the denominator \( 3+x \) is positive (since the numerator 2 is always positive). Therefore, we need: \[ 3+x > 0 \] Solving this inequality gives: \[ x > -3 \] Thus, for \( C \) to be a real number, \( x \) must be greater than \(-3\). In conclusion, the value(s) of \( x \) for which \( C \) is a real number is: \[ \boxed{(-3, \infty)} \]