To find the area bounded by the curves \(y = \cos x\) and \(y = \sin 2x\) over the interval \(\left[\frac{\pi}{6}, \frac{\pi}{2}\right]\), we will follow these steps:
### Step 1: Find the points of intersection
We need to find the points where the curves intersect by setting them equal to each other:
\[
\cos x = \sin 2x
\]
Using the identity \(\sin 2x = 2 \sin x \cos x\), we can rewrite the equation:
\[
\cos x = 2 \sin x \cos x
\]
Rearranging gives:
\[
\cos x - 2 \sin x \cos x = 0
\]
Factoring out \(\cos x\):
\[
\cos x (1 - 2 \sin x) = 0
\]
This gives us two cases:
1. \(\cos x = 0\)
2. \(1 - 2 \sin x = 0\)
From \(\cos x = 0\), we find \(x = \frac{\pi}{2}\) (within our interval).
From \(1 - 2 \sin x = 0\):
\[
2 \sin x = 1 \implies \sin x = \frac{1}{2}
\]
This gives \(x = \frac{\pi}{6}\) (within our interval).
Thus, the points of intersection are \(x = \frac{\pi}{6}\) and \(x = \frac{\pi}{2}\).
### Step 2: Determine which function is on top
To find the area between the curves, we need to determine which function is greater in the interval \(\left[\frac{\pi}{6}, \frac{\pi}{2}\right]\).
- At \(x = \frac{\pi}{6}\):
\[
y = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \quad y = \sin\left(2 \cdot \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}
\]
- At \(x = \frac{\pi}{4}\):
\[
y = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad y = \sin\left(\frac{\pi}{2}\right) = 1
\]
Since \(\sin 2x\) is greater than \(\cos x\) in the interval \(\left(\frac{\pi}{6}, \frac{\pi}{2}\right)\), we will integrate \(\sin 2x - \cos x\).
### Step 3: Set up the integral
The area \(A\) can be expressed as:
\[
A = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (\sin 2x - \cos x) \, dx
\]
### Step 4: Compute the integral
Now we compute the integral:
\[
A = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sin 2x \, dx - \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos x \, dx
\]
Calculating each integral separately:
1. For \(\int \sin 2x \, dx\):
\[
\int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C
\]
2. For \(\int \cos x \, dx\):
\[
\int \cos x \, dx = \sin x + C
\]
Now substituting the limits:
\[
A = \left[-\frac{1}{2} \cos 2x\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} - \left[\sin x\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}
\]
Calculating each part:
1. For \(-\frac{1}{2} \cos 2x\):
At \(x = \frac{\pi}{2}\):
\[
-\frac{1}{2} \cos(\pi) = -\frac{1}{2}(-1) = \frac{1}{2}
\]
At \(x = \frac{\pi}{6}\):
\[
-\frac{1}{2} \cos\left(\frac{\pi}{3}\right) = -\frac{1}{2} \cdot \frac{1}{2} = -\frac{1}{4}
\]
Thus:
\[
\left[-\frac{1}{2} \cos 2x\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} = \frac{1}{2} - \left(-\frac{1}{4}\right) = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}
\]
2. For \(\sin x\):
At \(x = \frac{\pi}{2}\):
\[
\sin\left(\frac{\pi}{2}\right) = 1
\]
At \(x = \frac{\pi}{6}\):
\[
\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}
\]
Thus:
\[
\left[\sin x\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} = 1 - \frac{1}{2} = \frac{1}{2}
\]
### Step 5: Combine results
Now, we combine the results:
\[
A = \frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4}
\]
### Final Answer
The area bounded by the curves \(y = \cos x\) and \(y = \sin 2x\) over the interval \(\left[\frac{\pi}{6}, \frac{\pi}{2}\right]\) is:
\[
\boxed{\frac{1}{4}}
\]
