Find the cardinality of the set \(C\), which is defined as: \[ C = \left\{x \mid \sin(4x) = \frac{1}{2}, for x \in \left(-9{\pi}, 3{\pi}\right)\right\} \]
Step-by-step Solution:
The given expression is: \[ C = \left\{ x \, \left\lvert \, \sin 4x = \frac{1}{2}, \, x \in (-9\pi, 3\pi) \right. \right\} \] This indicates the set of \( x \) values for which \( \sin 4x = \frac{1}{2} \) in the interval \( (-9\pi, 3\pi) \). To find the values of \( x \), we can start by solving the equation \( \sin 4x = \frac{1}{2} \). The general solutions for \( \sin \theta = \frac{1}{2} \) are: \[ \theta = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2n\pi \] Thus for \( \sin 4x = \frac{1}{2} \), we have: \[ 4x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad 4x = \frac{5\pi}{6} + 2n\pi \] Solving for \( x \), we get: \[ x = \frac{\pi}{24} + \frac{n\pi}{2} \quad \text{or} \quad x = \frac{5\pi}{24} + \frac{n\pi}{2} \] Next, we check for values of \( x \) in the interval \( (-9\pi, 3\pi) \): For \( x = \frac{\pi}{24} + \frac{n\pi}{2} \), and \( x = \frac{5\pi}{24} + \frac{n\pi}{2} \), the corresponding values of \( n \) are calculated. Since the solutions are periodic, there will be 48 values in the interval \( (-9\pi, 3\pi) \). Thus, the set \( C \) contains 48 solutions for \( x \).