A curve $C$ is given parametrically by the equations $x = 2\cos^3 \theta$, $y = 2\sin^3 \theta$ where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. Show that the normal at the point with parameter $\theta$ has equation
$y \sin \theta = x \cos \theta + 2\left(\sin^4 \theta - \cos^4 \theta\right).$

1. Find the cartesian equation of the normal at point $Q$ by substituting $\theta = \frac{\pi}{6}$ into the given generic normal equation.

Given the normal equation for any parameter $\theta$:
$y \sin \theta = x \cos \theta + 2(\sin^4 \theta - \cos^4 \theta)$
Substitute $\theta = \frac{\pi}{6}$ for point $Q$:
$y \sin\left(\frac{\pi}{6}\right) = x \cos\left(\frac{\pi}{6}\right) + 2\left(\sin^4\left(\frac{\pi}{6}\right) - \cos^4\left(\frac{\pi}{6}\right)\right)$
$y \left(\frac{1}{2}\right) = x \left(\frac{\sqrt{3}}{2}\right) + 2\left( \left(\frac{1}{2}\right)^4 - \left(\frac{\sqrt{3}}{2}\right)^4 \right)$
$y \left(\frac{1}{2}\right) = x \left(\frac{\sqrt{3}}{2}\right) + 2\left( \frac{1}{16} - \frac{9}{16} \right)$
$\frac{1}{2} y = \frac{\sqrt{3}}{2} x - 1$
Multiply by 2 to simplify:
$y = \sqrt{3}x - 2$

2. Substitute the parametric coordinates of $P$ into the normal equation to show the required relationship.

Since the normal intersects the curve $C$ again at $P$, the coordinates of $P$ at $\theta = p$, given by $x = 2\cos^3 p$ and $y = 2\sin^3 p$, must satisfy the normal's equation $y = \sqrt{3}x - 2$:
$2\sin^3 p = \sqrt{3}(2\cos^3 p) - 2$
Divide the entire equation by 2:
$\sin^3 p = \sqrt{3}\cos^3 p - 1$
Rearranging the terms yields the required equation:
$\sin^3 p - \sqrt{3}\cos^3 p + 1 = 0 \quad \text{(shown)}$

3. Use a graphing calculator to solve for $p$.

Using a graphing calculator to solve the equation $\sin^3 p - \sqrt{3}\cos^3 p + 1 = 0$ for the given domain $-\frac{\pi}{2} < p < \frac{\pi}{2}$:
We obtain the roots $p = \frac{\pi}{6}$ or $p \approx -0.74453$.
Since $p = \frac{\pi}{6}$ corresponds to the initial point $Q$, we reject it for $P$.
Thus, $p \approx -0.74453$.

4. Calculate the coordinates of $P$ using the found value of $p$.

Substitute $p = -0.74453$ back into the parametric equations:
$x = 2\cos^3(-0.74453) \approx 0.79515$
$y = 2\sin^3(-0.74453) \approx -0.62236$
The coordinates of $P$ are $(0.795, -0.622)$ correct to 3 significant figures.