sin>1/2

To solve the inequality \(\sin(x) > \frac{1}{2}\), we need to determine the values of \(x\) for which the sine function exceeds \(\frac{1}{2}\).

The sine function, \(\sin(x)\), is periodic with a period of \(2\pi\). The specific angle where \(\sin(x) = \frac{1}{2}\) is at:

• \(x = \frac{\pi}{6} + 2k\pi\) and
• \(x = \frac{5\pi}{6} + 2k\pi\)

for any integer \(k\).

To find where \(\sin(x) > \frac{1}{2}\), we observe that:

1. \(\sin(x)\) is greater than \(\frac{1}{2}\) between the angles:
   • \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) in each interval.

Thus, the solution to the inequality \(\sin(x) > \frac{1}{2}\) is:

\[ x \in \left(\frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi\right) \quad \text{for any integer } k. \]

In summary, the solution set for \(\sin(x) > \frac{1}{2}\) includes all \(x\) values in the open intervals defined above for every integer \(k\).