Find the most general value of ` theta ` which satisfies the equation ` sin theta =(1)/(2) and tan theta =(1)/(sqrt(3))`

Answer

To solve the equations \( \sin \theta = \frac{1}{2} \) and \( \tan \theta = \frac{1}{\sqrt{3}} \), we will find the most general value of \( \theta \) that satisfies both conditions.

### Step 1: Analyze the equations
We have two equations:
1. \( \sin \theta = \frac{1}{2} \)
2. \( \tan \theta = \frac{1}{\sqrt{3}} \)

### Step 2: Find the angles for \( \sin \theta = \frac{1}{2} \)
The sine function equals \( \frac{1}{2} \) at:
- \( \theta = \frac{\pi}{6} \) (in the first quadrant)
- \( \theta = \frac{5\pi}{6} \) (in the second quadrant)

### Step 3: Find the angles for \( \tan \theta = \frac{1}{\sqrt{3}} \)
The tangent function equals \( \frac{1}{\sqrt{3}} \) at:
- \( \theta = \frac{\pi}{6} \) (in the first quadrant)
- \( \theta = \frac{7\pi}{6} \) (in the third quadrant)

### Step 4: Determine the common angles
Both equations yield \( \theta = \frac{\pi}{6} \) as a common solution.

### Step 5: Consider the periodic nature of the functions
The sine function has a period of \( 2\pi \), and the tangent function also has a period of \( \pi \). Therefore, the general solutions can be expressed as:
- For sine: \( \theta = \frac{\pi}{6} + 2n\pi \) where \( n \in \mathbb{Z} \)
- For tangent: \( \theta = \frac{\pi}{6} + k\pi \) where \( k \in \mathbb{Z} \)

### Step 6: Combine the general solutions
Since both sine and tangent must be satisfied, we take the solution from sine and add the periodicity of tangent:
- The most general value of \( \theta \) satisfying both conditions is:
\[ \theta = \frac{\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]

### Final Answer
Thus, the most general value of \( \theta \) that satisfies both equations is:
\[ \theta = \frac{\pi}{6} + n\pi \quad (n \in \mathbb{Z}) \]

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