已知 $f(x)$ 满足 $f'(x) = \cos 4x - \sin 2x$。又已知 $f\left(\frac{\pi}{2}\right) = \frac{1}{4}$，证明 $f''(x) + 4f(x) = 3 - 3\sin 4x$。

1. Find the second derivative $f''(x)$ by differentiating $f'(x)$.

$f''(x) = \frac{d}{dx}(\cos 4x - \sin 2x)$
$f''(x) = -4\sin 4x - 2\cos 2x$

2. Find the general form of $f(x)$ by integrating $f'(x)$.

$f(x) = \int (\cos 4x - \sin 2x) \, dx$
$f(x) = \frac{1}{4}\sin 4x + \frac{1}{2}\cos 2x + C$

3. Use the given condition $f\left(\frac{\pi}{2}\right) = \frac{1}{4}$ to find the constant of integration $C$.

$f\left(\frac{\pi}{2}\right) = \frac{1}{4}\sin\left(4\left(\frac{\pi}{2}\right)\right) + \frac{1}{2}\cos\left(2\left(\frac{\pi}{2}\right)\right) + C = \frac{1}{4}$
$\frac{1}{4}\sin(2\pi) + \frac{1}{2}\cos(\pi) + C = \frac{1}{4}$
$0 + \frac{1}{2}(-1) + C = \frac{1}{4}$
$-\frac{1}{2} + C = \frac{1}{4}$
$C = \frac{3}{4}$
So, $f(x) = \frac{1}{4}\sin 4x + \frac{1}{2}\cos 2x + \frac{3}{4}$

4. Substitute $f''(x)$ and $f(x)$ into the expression $f''(x) + 4f(x)$ to show the required result.

$f''(x) + 4f(x) = (-4\sin 4x - 2\cos 2x) + 4\left(\frac{1}{4}\sin 4x + \frac{1}{2}\cos 2x + \frac{3}{4}\right)$
$= -4\sin 4x - 2\cos 2x + \sin 4x + 2\cos 2x + 3$
$= -3\sin 4x + 3$
$= 3 - 3\sin 4x \quad \text{(shown)}$