Solve the trigonometric equation for all values \( 0 \leq x<2 \pi \) \[ \sec x-2=0 \]

To solve the equation \( \sec x - 2 = 0 \), we first isolate the secant function: \[ \sec x = 2 \] Since \(\sec x\) is the reciprocal of \(\cos x\), we can rewrite this as: \[ \frac{1}{\cos x} = 2 \] Taking the reciprocal of both sides gives us: \[ \cos x = \frac{1}{2} \] Now, we need to find the values of \(x\) in the interval \(0 \leq x < 2\pi\) where \(\cos x = \frac{1}{2}\). The cosine function is equal to \(\frac{1}{2}\) at: \[ x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3} \] Thus, the solutions to the equation \( \sec x - 2 = 0 \) in the interval \(0 \leq x < 2\pi\) are: \[ x = \frac{\pi}{3}, \quad x = \frac{5\pi}{3} \]