The initial value problem presented is:

$\cos(x)y' + \sin(x)y = 2\cos^3(x)\sin(x) - 1$ with the condition $y\left(\frac{\pi}{4}\right) = 3\sqrt{2}.$

We will solve this first-order linear differential equation using an integrating factor approach. Let's proceed step by step.

Step 1: Rewrite the equation in standard linear form

The equation is already in the form:

$y' + P(x)y = Q(x)$ where $P(x) = \frac{\sin(x)}{\cos(x)} = \tan(x)$ and $Q(x) = \frac{2\cos^3(x)\sin(x) - 1}{\cos(x)}$.

Thus, the equation becomes:

$y' + \tan(x) y = 2\cos^2(x)\sin(x) - \sec(x).$

Step 2: Find the integrating factor

The integrating factor $\mu(x)$ is given by:

$\mu(x) = e^{\int P(x) dx} = e^{\int \tan(x) dx}.$

The integral of $\tan(x)$ is $\ln|\sec(x)|$, so the integrating factor becomes:

$\mu(x) = e^{\ln|\sec(x)|} = \sec(x).$

Step 3: Multiply the entire equation by the integrating factor

Multiplying both sides of the equation by $\sec(x)$, we get:

$\sec(x)y' + \sec(x)\tan(x)y = 2\cos^2(x)\sin(x) - \sec(x).$

The left side simplifies to:

$\frac{d}{dx} \left( y \sec(x) \right).$

Thus, the equation becomes:

[ \frac{d}{dx} \left( y \sec(x) \right) = 2\cos^2(x)\sin(x)