Find the area enclosed with in the curve
`y=sin 2x , y=sqrt(3) sin x , x=0 ,x=(pi)/(6)`

Answer

To find the area enclosed within the curves \( y = \sin 2x \), \( y = \sqrt{3} \sin x \), and the vertical lines \( x = 0 \) and \( x = \frac{\pi}{6} \), we will follow these steps:

### Step 1: Set the equations equal to find intersection points
We start by equating the two functions:
\[ \sin 2x = \sqrt{3} \sin x \]
Using the double angle identity, we can rewrite \( \sin 2x \):
\[ 2 \sin x \cos x = \sqrt{3} \sin x \]
Assuming \( \sin x \neq 0 \), we can divide both sides by \( \sin x \):
\[ 2 \cos x = \sqrt{3} \]
Now, solve for \( \cos x \):
\[ \cos x = \frac{\sqrt{3}}{2} \]

### Step 2: Find the value of \( x \)
The value of \( x \) that satisfies \( \cos x = \frac{\sqrt{3}}{2} \) is:
\[ x = \frac{\pi}{6} \]

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{\pi}{6} \) is given by:
\[ A = \int_{0}^{\frac{\pi}{6}} (\sin 2x - \sqrt{3} \sin x) \, dx \]

### Step 4: Integrate the functions
Now we will compute the integral:
\[ A = \int_{0}^{\frac{\pi}{6}} \sin 2x \, dx - \int_{0}^{\frac{\pi}{6}} \sqrt{3} \sin x \, dx \]

Calculating the first integral:
\[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x \]
Thus,
\[ \int_{0}^{\frac{\pi}{6}} \sin 2x \, dx = \left[-\frac{1}{2} \cos 2x \right]_{0}^{\frac{\pi}{6}} = -\frac{1}{2} \left( \cos\left(\frac{\pi}{3}\right) - \cos(0) \right) = -\frac{1}{2} \left( \frac{1}{2} - 1 \right) = -\frac{1}{2} \left( -\frac{1}{2} \right) = \frac{1}{4} \]

Calculating the second integral:
\[ \int \sqrt{3} \sin x \, dx = -\sqrt{3} \cos x \]
Thus,
\[ \int_{0}^{\frac{\pi}{6}} \sqrt{3} \sin x \, dx = \left[-\sqrt{3} \cos x \right]_{0}^{\frac{\pi}{6}} = -\sqrt{3} \left( \cos\left(\frac{\pi}{6}\right) - \cos(0) \right) = -\sqrt{3} \left( \frac{\sqrt{3}}{2} - 1 \right) = -\sqrt{3} \left( \frac{\sqrt{3}}{2} - \frac{2}{2} \right) = -\sqrt{3} \left( \frac{\sqrt{3} - 2}{2} \right) = \frac{3 - 2\sqrt{3}}{2} \]

### Step 5: Combine the results
Now, substituting back into the area formula:
\[ A = \frac{1}{4} - \left( \frac{3 - 2\sqrt{3}}{2} \right) \]
Simplifying this gives:
\[ A = \frac{1}{4} - \frac{3}{2} + \sqrt{3} = \frac{1}{4} - \frac{6}{4} + \frac{2\sqrt{3}}{4} = \frac{1 - 6 + 2\sqrt{3}}{4} = \frac{-5 + 2\sqrt{3}}{4} \]

### Final Answer
Thus, the area enclosed is:
\[ A = \frac{7}{4} - \sqrt{3} \]