To simplify the expression \( A = \frac{\sin 3x \cos 2x + \sin x \cos 6x}{\sin 4x} \), we can utilize trigonometric identities.

### Step 1: Apply Trigonometric Identities

We will use the product-to-sum identities:
\[
\sin A \cos B = \frac{1}{2} (\sin(A + B) + \sin(A - B))
\]

#### For the first term: \(\sin 3x \cos 2x\)
\[
\sin 3x \cos 2x = \frac{1}{2} (\sin(3x + 2x) + \sin(3x - 2x)) = \frac{1}{2} (\sin 5x + \sin x)
\]

#### For the second term: \(\sin x \cos 6x\)
\[
\sin x \cos 6x = \frac{1}{2} (\sin(6x + x) + \sin(6x - x)) = \frac{1}{2} (\sin 7x + \sin 5x)
\]

### Step 2: Substitute back into A

Now substitute these results back into \( A \):

\[
A = \frac{\frac{1}{2} (\sin 5x + \sin x) + \frac{1}{2} (\sin 7x + \sin 5x)}{\sin 4x}
\]

Combining the terms in the numerator:

\[
A = \frac{\frac{1}{2} (2 \sin 5x + \sin x + \sin 7x)}{\sin 4x} = \frac{\sin 5x + \frac{1}{2} (\sin x + \sin 7x)}{\sin 4x}
\]

### Step 3: Simplify Further

Now we evaluate \( \sin x + \sin 7x \):
Using the sum-to-product identity:
\[
\sin x + \sin 7x = 2 \sin\left(\frac{x + 7x}{2}\right) \cos\left(\frac{7x - x}{2}\right) = 2 \sin(4x) \cos(3x)
\]

### Step 4: Substitute and Simplify Again

Substituting this back into \( A \):

\[
A = \frac{\sin 5x + \sin(4x) \cos(3x)}{\sin 4x}
\]

You can separate the terms:

\[
A = \frac{\sin 5x}{\sin 4x} + \frac{\sin(4x) \cos(3x)}{\sin 4x} = \frac{\sin 5x}{\sin 4x} + \cos(3x)
\]

### Final Result

Thus, the simplified form of the expression \( A \) is:

\[
A = \frac{\sin 5x}{\sin 4x} + \cos(3x)
\]