The graphs of \( f(x) = \cos x \) and \( g(x) = \sin 2x \) intersect at points A, B, and C. Given B at \( \left( \frac{\pi}{2}, 0 \right) \):

\[ f\left( \frac{\pi}{2} \right) = \cos \frac{\pi}{2} = 0, \quad g\left( \frac{\pi}{2} \right) = \sin \left( 2 \cdot \frac{\pi}{2} \right) = \sin \pi = 0 \]

Confirming B is an intersection point. To find A and C, set \( f(x) = g(x) \):

\[ \cos x = \sin 2x \]

Use the identity \( \sin 2x = 2 \sin x \cos x \):

\[ \cos x = 2 \sin x \cos x \]

\[ \cos x – 2 \sin x \cos x = 0 \]

\[ \cos x (1 – 2 \sin x) = 0 \]

Solutions:

1. \( \cos x = 0 \):

\[ x = \frac{\pi}{2} \quad (\text{point B}) \]

2. \( 1 – 2 \sin x = 0 \):

\[ 2 \sin x = 1 \]

\[ \sin x = \frac{1}{2} \]

In \( [0, \pi] \):

\[ x = \frac{\pi}{6} \quad (\text{since } \sin \frac{\pi}{6} = \frac{1}{2}) \]

\[ x = \pi – \frac{\pi}{6} = \frac{5\pi}{6} \quad (\text{since } \sin (\pi – x) = \sin x) \]

From the graph, point A has the smallest x-coordinate, and point C the largest:

– Point A: \( x = \frac{\pi}{6} \)

– Point C: \( x = \frac{5\pi}{6} \)

Verify:

At \( x = \frac{\pi}{6} \):

\[ f\left( \frac{\pi}{6} \right) = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}, \quad g\left( \frac{\pi}{6} \right) = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \]

At \( x = \frac{5\pi}{6} \):

\[ f\left( \frac{5\pi}{6} \right) = \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}, \quad g\left( \frac{5\pi}{6} \right) = \sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2} \]

Answer: x-coordinate of A: \( \frac{\pi}{6} \), x-coordinate of C: \( \frac{5\pi}{6} \).

Region \( R \) is enclosed by \( f(x) = \cos x \) and \( g(x) = \sin 2x \) from \( x = \frac{\pi}{2} \) to \( x = \frac{5\pi}{6} \), where \( g(x) \geq f(x) \).

The area is:

\[ \text{Area} = \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} (g(x) – f(x)) \, dx = \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} (\sin 2x – \cos x) \, dx \]

Method 1:

Integrate \( \cos x – \sin 2x \):

\[ \int (\cos x – \sin 2x) \, dx \]

\[ \int \cos x \, dx = \sin x \]

\[ \int -\sin 2x \, dx = -\int \sin 2x \, dx \]

Substitute \( u = 2x \), \( du = 2 \, dx \), \( dx = \frac{du}{2} \):

\[ -\int \sin 2x \, dx = -\int \sin u \cdot \frac{du}{2} = -\frac{1}{2} (-\cos u) = \frac{1}{2} \cos 2x \]

Antiderivative:

\[ \sin x + \frac{1}{2} \cos 2x \]

Evaluate:

\[ \left[ \sin x + \frac{1}{2} \cos 2x \right]_{\frac{\pi}{2}}^{\frac{5\pi}{6}} \]

At \( x = \frac{5\pi}{6} \):

\[ \sin \frac{5\pi}{6} = \frac{1}{2}, \quad \cos \left( 2 \cdot \frac{5\pi}{6} \right) = \cos \frac{5\pi}{3} = \frac{1}{2} \]

\[ \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \]

At \( x = \frac{\pi}{2} \):

\[ \sin \frac{\pi}{2} = 1, \quad \cos \left( 2 \cdot \frac{\pi}{2} \right) = \cos \pi = -1 \]

\[ 1 + \frac{1}{2} \cdot (-1) = 1 – \frac{1}{2} = \frac{1}{2} \]

\[ \frac{3}{4} – \frac{1}{2} = \frac{3}{4} – \frac{2}{4} = \frac{1}{4} \]

Area: \( \frac{1}{4} \).

Method 2:

Compute separately:

\[ \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} \cos x \, dx = \left[ \sin x \right]_{\frac{\pi}{2}}^{\frac{5\pi}{6}} = \sin \frac{5\pi}{6} – \sin \frac{\pi}{2} = \frac{1}{2} – 1 = -\frac{1}{2} \]

\[ \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} \sin 2x \, dx = \left[ -\frac{1}{2} \cos 2x \right]_{\frac{\pi}{2}}^{\frac{5\pi}{6}} \]

\[ = -\frac{1}{2} \cos \frac{5\pi}{3} + \frac{1}{2} \cos \pi = -\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot (-1) = -\frac{1}{4} – \frac{1}{2} = -\frac{3}{4} \]

Area:

\[ \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} (\sin 2x – \cos x) \, dx = -\frac{3}{4} – \left(-\frac{1}{2}\right) = -\frac{3}{4} + \frac{1}{2} = -\frac{1}{4} \]

Take absolute value: \( \left| -\frac{1}{4} \right| = \frac{1}{4} \).

Answer: Area of \( R \) is \( \frac{1}{4} \).