The problem asks to simplify the trigonometric expression $\sin 7x + \sin x - 2\sin 2x \cos 3x$.

1. Problem Description

The problem asks to simplify the trigonometric expression $\sin 7x + \sin x - 2\sin 2x \cos 3x$.

2. Solution Steps

We can use the sum-to-product formula for $\sin A + \sin B$, which is:

$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$

Applying this formula to $\sin 7x + \sin x$, we have:

$\sin 7x + \sin x = 2 \sin\left(\frac{7x+x}{2}\right) \cos\left(\frac{7x-x}{2}\right) = 2 \sin\left(\frac{8x}{2}\right) \cos\left(\frac{6x}{2}\right) = 2 \sin 4x \cos 3x$

Thus, the given expression becomes:

$2 \sin 4x \cos 3x - 2 \sin 2x \cos 3x$

We can factor out $2\cos 3x$:

$2 \cos 3x (\sin 4x - \sin 2x)$

Now we use the sum-to-product formula for $\sin A - \sin B$:

$\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$

Applying this formula to $\sin 4x - \sin 2x$, we have:

$\sin 4x - \sin 2x = 2 \cos\left(\frac{4x+2x}{2}\right) \sin\left(\frac{4x-2x}{2}\right) = 2 \cos\left(\frac{6x}{2}\right) \sin\left(\frac{2x}{2}\right) = 2 \cos 3x \sin x$

Substituting this back into our expression:

$2 \cos 3x (2 \cos 3x \sin x) = 4 \cos^2 3x \sin x$

3. Final Answer

The simplified expression is $4 \cos^2 3x \sin x$. Therefore, the answer is option c.