6 Solve for \( [0,2 \pi] \) : a \( \cos 2\left(x-\frac{\pi}{2}\right)=\frac{1}{2} \)

To solve the equation \( \cos 2\left(x - \frac{\pi}{2}\right) = \frac{1}{2} \), we start by recognizing that \( \cos \theta = \frac{1}{2} \) leads to two general solutions: 1. \( \theta = \frac{\pi}{3} + 2k\pi \) 2. \( \theta = \frac{5\pi}{3} + 2k\pi \) Here, \( k \) is any integer. Substituting \( \theta = 2\left(x - \frac{\pi}{2}\right) \) gives: 1. \( 2\left(x - \frac{\pi}{2}\right) = \frac{\pi}{3} + 2k\pi \) 2. \( 2\left(x - \frac{\pi}{2}\right) = \frac{5\pi}{3} + 2k\pi \) Now, let's simplify each case: For the first equation: \[ 2x - \pi = \frac{\pi}{3} + 2k\pi \implies 2x = \frac{\pi}{3} + \pi + 2k\pi \implies 2x = \frac{4\pi}{3} + 2k\pi \implies x = \frac{2\pi}{3} + k\pi \] For the second equation: \[ 2x - \pi = \frac{5\pi}{3} + 2k\pi \implies 2x = \frac{5\pi}{3} + \pi + 2k\pi \implies 2x = \frac{8\pi}{3} + 2k\pi \implies x = \frac{4\pi}{3} + k\pi \] Now we will check the values of \( x \) for \( k = 0 \) (since we are constraining \( x \) to the interval \([0, 2\pi]\)): 1. From the first equation: - \( x = \frac{2\pi}{3} + 0 \cdot \pi = \frac{2\pi}{3} \) (valid) 2. From the first equation: - \( x = \frac{2\pi}{3} + 1 \cdot \pi = \frac{2\pi}{3} + \pi = \frac{5\pi}{3} \) (valid) 3. From the second equation: - \( x = \frac{4\pi}{3} + 0 \cdot \pi = \frac{4\pi}{3} \) (valid) 4. From the second equation: - \( x = \frac{4\pi}{3} + 1 \cdot \pi = \frac{4\pi}{3} + \pi = \frac{7\pi}{3} \) (not valid, as \( \frac{7\pi}{3} > 2\pi \)) So, the solutions to the equation \( \cos 2\left(x - \frac{\pi}{2}\right) = \frac{1}{2} \) in the interval \([0, 2\pi]\) are: \[ x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \]