\( 1-2\sin (x)>0 \)

To solve the inequality \( 1 - 2\sin(x) > 0 \), we start by isolating \(\sin(x)\): \[ 1 > 2\sin(x) \implies \sin(x) < \frac{1}{2} \] This tells us that we're looking for the angles \(x\) for which the sine function is less than \(\frac{1}{2}\). In the unit circle, \(\sin(x) = \frac{1}{2}\) at \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\) within one full rotation (0 to \(2\pi\)). Therefore, the solution set will extend from \(x < \frac{\pi}{6}\) and from \(x > \frac{5\pi}{6}\). However, because the sine function is periodic, we also need to consider the general solutions: \[ x < \frac{\pi}{6} + 2k\pi \quad \text{and} \quad x > \frac{5\pi}{6} + 2k\pi \quad \text{for any integer } k. \] Make sure to test values in the intervals to ensure correctness! For instance, \(x = 0\) gives \(\sin(0) = 0\), which satisfies the inequality. In contrast, \(x = \frac{\pi}{2}\) gives \(\sin\left(\frac{\pi}{2}\right) = 1\), violating the condition. Keep those critical points in mind as you analyze the solutions!