To find the area of the region enclosed by the curves \( y = \cos x \), \( y = \sin 2x \), \( x = 0 \), and \( x = \frac{\pi}{2} \), we will follow these steps:
### Step 1: Find the Points of Intersection
We need to find where the curves \( y = \cos x \) and \( y = \sin 2x \) intersect. This requires solving the equation:
\[
\cos x = \sin 2x
\]
Using the identity \( \sin 2x = 2 \sin x \cos x \), we can rewrite the equation as:
\[
\cos x = 2 \sin x \cos x
\]
Assuming \( \cos x \neq 0 \), we can divide both sides by \( \cos x \):
\[
1 = 2 \sin x
\]
This simplifies to:
\[
\sin x = \frac{1}{2}
\]
The solutions for this equation in the interval \( [0, \frac{\pi}{2}] \) is:
\[
x = \frac{\pi}{6}
\]
### Step 2: Set Up the Integral for the Area
The area \( A \) between the curves from \( x = 0 \) to \( x = \frac{\pi}{2} \) can be found by calculating the area from \( 0 \) to \( \frac{\pi}{6} \) and from \( \frac{\pi}{6} \) to \( \frac{\pi}{2} \):
\[
A = \int_0^{\frac{\pi}{6}} (\cos x - \sin 2x) \, dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (\sin 2x - \cos x) \, dx
\]
### Step 3: Evaluate the First Integral
Calculating the first integral:
\[
\int_0^{\frac{\pi}{6}} (\cos x - \sin 2x) \, dx
\]
This can be split into two parts:
\[
\int_0^{\frac{\pi}{6}} \cos x \, dx - \int_0^{\frac{\pi}{6}} \sin 2x \, dx
\]
Calculating these integrals:
1. \( \int \cos x \, dx = \sin x \)
2. \( \int \sin 2x \, dx = -\frac{1}{2} \cos 2x \)
Evaluating:
\[
\sin x \Big|_0^{\frac{\pi}{6}} - \left(-\frac{1}{2} \cos 2x \Big|_0^{\frac{\pi}{6}}\right)
\]
Calculating:
\[
\sin\left(\frac{\pi}{6}\right) - \sin(0) + \frac{1}{2} \left( \cos(0) - \cos\left(\frac{\pi}{3}\right) \right)
\]
\[
= \frac{1}{2} + \frac{1}{2} \left( 1 - \frac{1}{2} \right) = \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}
\]
### Step 4: Evaluate the Second Integral
Now for the second integral:
\[
\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (\sin 2x - \cos x) \, dx
\]
This can be split similarly:
\[
\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sin 2x \, dx - \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos x \, dx
\]
Calculating:
\[
-\frac{1}{2} \cos 2x \Big|_{\frac{\pi}{6}}^{\frac{\pi}{2}} - \sin x \Big|_{\frac{\pi}{6}}^{\frac{\pi}{2}}
\]
Evaluating:
\[
-\frac{1}{2} \left( \cos(\pi) - \cos\left(\frac{\pi}{3}\right) \right) - \left( \sin\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{6}\right) \right)
\]
\[
= -\frac{1}{2} \left( -1 - \frac{1}{2} \right) - \left( 1 - \frac{1}{2} \right)
\]
\[
= -\frac{1}{2} \left( -\frac{3}{2} \right) - \frac{1}{2} = \frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4}
\]
### Step 5: Combine the Areas
Now, we add the areas from both integrals:
\[
A = \frac{3}{4} + \frac{1}{4} = 1
\]
### Final Answer
Thus, the area of the region enclosed by the curves is:
\[
\boxed{1}
\]
