Determine the value(s) of \( x \) for which \( f(x) \) is discontinuous if: \( f(x)=\frac{1}{2-4 \cos 2 x} ; x \in[0 ; 2 \pi] \)

To determine the values of \( x \) for which the function \( f(x) = \frac{1}{2 - 4 \cos 2x} \) is discontinuous, we need to find where the denominator equals zero, as division by zero leads to discontinuity. Setting the denominator equal to zero gives: \[ 2 - 4 \cos 2x = 0 \] Solving for \( \cos 2x \): \[ 4 \cos 2x = 2 \\ \cos 2x = \frac{1}{2} \] The cosine function equals \( \frac{1}{2} \) at specific angles. The general solutions for \( 2x \) where \( \cos 2x = \frac{1}{2} \) are: \[ 2x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad 2x = \frac{5\pi}{3} + 2k\pi \quad (k \in \mathbb{Z}) \] Dividing through by 2 gives: \[ x = \frac{\pi}{6} + k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + k\pi \] Now, we restrict \( x \) to the interval \([0, 2\pi]\). Finding values for \( k \): 1. For \( x = \frac{\pi}{6} + k\pi \): - If \( k = 0 \), \( x = \frac{\pi}{6} \) - If \( k = 1 \), \( x = \frac{7\pi}{6} \) 2. For \( x = \frac{5\pi}{6} + k\pi \): - If \( k = 0 \), \( x = \frac{5\pi}{6} \) - If \( k = 1 \), \( x = \frac{11\pi}{6} \) So the discontinuous points in the interval \([0, 2\pi]\) are: \[ x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \] In conclusion, the function \( f(x) \) is discontinuous at \( x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \).