Simplify Trigonometric Expression

The question asks for the value of the trigonometric expression $\frac{\cos 3x + \cos x}{\sin 3x - \sin x}$. We will use sum-to-product and difference-to-product trigonometric identities to simplify this expression.

Apply Trigonometric Identities

Use the sum-to-product identity for the numerator:

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

For the numerator, $A=3x$ and $B=x$:

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

Use the difference-to-product identity for the denominator:

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

For the denominator, $A=3x$ and $B=x$:

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

Simplify the Expression

Now, substitute the simplified numerator and denominator back into the original expression:

$\frac{\cos 3x + \cos x}{\sin 3x - \sin x} = \frac{2 \cos(2x) \cos(x)}{2 \cos(2x) \sin(x)}$

Cancel out the common term $2 \cos(2x)$ (assuming $\cos(2x) \neq 0$):

$\frac{\cos(x)}{\sin(x)}$

Final Result

The expression simplifies to $\cot(x)$.

Final Answer: The final answer is $\boxed{\cot x}$