We are asked to simplify the expression:

$4\cos(4x) - 5\cos(2x) + \sin(x) \sin(3x) + 3$

Step 1: Simplify the trigonometric terms

Use product-to-sum identities

For the product $\sin(x) \sin(3x)$, we can use the product-to-sum identity:

$\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$

Substituting $A = x$ and $B = 3x$, we get:

[ \sin(x) \sin(3x) = \frac{1}{2} [\cos(x - 3x) - \cos(x + 3x)] = \frac{1}{2} [\cos(-2x) - \cos(4x)] ]

Since $\cos(-2x) = \cos(2x)$, this simplifies to:

$\sin(x) \sin(3x) = \frac{1}{2} [\cos(2x) - \cos(4x)]$

Step 2: Substitute into the original expression

Now, substitute this result back into the original expression:

$4\cos(4x) - 5\cos(2x) + \frac{1}{2} [\cos(2x) - \cos(4x)] + 3$

Step 3: Combine like terms

Group the $\cos(4x)$ and $\cos(2x)$ terms:

$(4\cos(4x) - \frac{1}{2} \cos(4x)) + (-5\cos(2x) + \frac{1}{2} \cos(2x)) + 3$

Simplify each group:

$(4 - \frac{1}{2}) \cos(4x) = \frac{7}{2} \cos(4x)$ $(-5 + \frac{1}{2}) \cos(2x) = -\frac{9}{2} \cos(2x)$

Thus, the expression simplifies to:

$\frac{7}{2} \cos(4x) - \frac{9}{2} \cos(2x) + 3$

This is the simplified form of the expression.

Would you like more details, or do you have any further questions?

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Here are some related questions you can explore:

1. How do you apply sum and difference formulas for sine and cosine?
2. What are product-to-sum identities, and how are they derived?
3. Can this expression be solved for a specific value of $x$?
4. What are the general solutions for trigonometric equations involving $\cos(4x)$ and $\cos(2x)$?
5. How do phase shifts affect trigonometric functions?

Tip: When working with trigonometric identities, it's often helpful to rewrite products as sums to simplify expressions.