A particular curve for which $\frac{d^2y}{dx^2} = \sin 3x - \cos \frac{1}{2}x$ passes through the point $Q\left(\frac{\pi}{3}, 2\sqrt{3}\right)$. The gradient of the curve at $Q$ is 1. Find the equation of the curve.

1. Find the expression for $\frac{dy}{dx}$ by integrating $\frac{d^2y}{dx^2}$.

$\frac{dy}{dx} = \int \left(\sin 3x - \cos \frac{1}{2}x\right) dx$
$\frac{dy}{dx} = -\frac{1}{3}\cos 3x - 2\sin \frac{1}{2}x + c$

2. Use the given gradient at $Q$ to find the constant of integration $c$.

At $x = \frac{\pi}{3}$, $\frac{dy}{dx} = 1$:
$1 = -\frac{1}{3}\cos\left(3\left(\frac{\pi}{3}\right)\right) - 2\sin\left(\frac{1}{2}\left(\frac{\pi}{3}\right)\right) + c$
$1 = -\frac{1}{3}\cos(\pi) - 2\sin\left(\frac{\pi}{6}\right) + c$
$1 = -\frac{1}{3}(-1) - 2\left(\frac{1}{2}\right) + c$
$1 = \frac{1}{3} - 1 + c$
$1 = -\frac{2}{3} + c \implies c = \frac{5}{3}$
So, $\frac{dy}{dx} = -\frac{1}{3}\cos 3x - 2\sin \frac{1}{2}x + \frac{5}{3}$

3. Find the expression for $y$ by integrating $\frac{dy}{dx}$.

$y = \int \left(-\frac{1}{3}\cos 3x - 2\sin \frac{1}{2}x + \frac{5}{3}\right) dx$
$y = -\frac{1}{9}\sin 3x + 4\cos \frac{1}{2}x + \frac{5}{3}x + d$

4. Use the coordinates of $Q$ to find the constant of integration $d$.

At $x = \frac{\pi}{3}$, $y = 2\sqrt{3}$:
$2\sqrt{3} = -\frac{1}{9}\sin\left(3\left(\frac{\pi}{3}\right)\right) + 4\cos\left(\frac{1}{2}\left(\frac{\pi}{3}\right)\right) + \frac{5}{3}\left(\frac{\pi}{3}\right) + d$
$2\sqrt{3} = -\frac{1}{9}\sin(\pi) + 4\cos\left(\frac{\pi}{6}\right) + \frac{5\pi}{9} + d$
$2\sqrt{3} = 0 + 4\left(\frac{\sqrt{3}}{2}\right) + \frac{5\pi}{9} + d$
$2\sqrt{3} = 2\sqrt{3} + \frac{5\pi}{9} + d$
$d = -\frac{5\pi}{9}$

Equation of the curve:
$y = -\frac{1}{9}\sin 3x + 4\cos \frac{1}{2}x + \frac{5}{3}x - \frac{5\pi}{9}$