Find the solution set of the inequality `sinx > 1/2.`

Answer

To solve the inequality \( \sin x > \frac{1}{2} \), we will follow these steps:

### Step 1: Identify the critical points
We know that \( \sin x = \frac{1}{2} \) at specific angles. The primary angles where this occurs are:
- \( x = \frac{\pi}{6} \) (in the first quadrant)
- \( x = \frac{5\pi}{6} \) (in the second quadrant)

### Step 2: Analyze the sine function
The sine function is positive in the first and second quadrants. Therefore, we need to determine the intervals where \( \sin x \) is greater than \( \frac{1}{2} \).

### Step 3: Determine the intervals
From the unit circle and the behavior of the sine function:
- Between \( \frac{\pi}{6} \) and \( \frac{5\pi}{6} \), the sine function is greater than \( \frac{1}{2} \).

### Step 4: Write the solution set
The solution set for the inequality \( \sin x > \frac{1}{2} \) can be expressed in interval notation as:
\[ x \in \left( \frac{\pi}{6}, \frac{5\pi}{6} \right) \]

### Step 5: Consider periodicity
Since the sine function is periodic with a period of \( 2\pi \), we can express the general solution as:
\[ x \in \left( \frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi \right) \quad \text{for } k \in \mathbb{Z} \]

### Final Answer
Thus, the complete solution set for the inequality \( \sin x > \frac{1}{2} \) is:
\[ \bigcup_{k \in \mathbb{Z}} \left( \frac{\pi}{6} + 2k\pi, \frac{5\pi}{6} + 2k\pi \right) \]