Step 1 :The given equation is \(\cos 2x = \frac{1}{2}\). To solve this equation, we need to find the values of \(x\) in the interval \([0, 2\pi)\) that satisfy this equation.
Step 2 :The cosine function \(\cos \theta\) equals \(\frac{1}{2}\) at \(\theta = \frac{\pi}{3}\) and \(\theta = \frac{5\pi}{3}\) in the interval \([0, 2\pi)\).
Step 3 :Since the equation is \(\cos 2x = \frac{1}{2}\), we set \(2x = \frac{\pi}{3}\) and \(2x = \frac{5\pi}{3}\), and solve for \(x\).
Step 4 :The solutions to the equations \(2x = \frac{\pi}{3}\) and \(2x = \frac{5\pi}{3}\) are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\) respectively.
Step 5 :However, since the original equation is \(\cos 2x = \frac{1}{2}\), we need to consider the periodicity of the cosine function. The cosine function has a period of \(2\pi\), so we need to add multiples of \(\pi\) to our solutions until we reach \(2\pi\).
Step 6 :The solutions to the equation \(\cos 2x = \frac{1}{2}\) in the interval \([0, 2\pi)\) are \(x = \frac{\pi}{6}\), \(x = \frac{7\pi}{6}\), \(x = \frac{5\pi}{6}\), and \(x = \frac{11\pi}{6}\).
Step 7 :Final Answer: \(\boxed{x = \frac{\pi}{6}, \frac{7\pi}{6}, \frac{5\pi}{6}, \frac{11\pi}{6}}\)