A and B are acute angles such that $\sin(A - B) = \frac{3}{8}$ and $\sin A \cos B = \frac{5}{8}$. Without using a calculator, find the value of $\cos A \sin B$. Express $2\sin 2\theta(\sec \theta - \tan \theta)$ as a quadratic expression in $\sin \theta$. Use your answer to part (b) to find, for $0 \le \theta \le 2\pi$, the exact solutions of the equation $2\sin 2\theta(\sec \theta - \tan \theta) + 3 = 0$.

1. State the compound angle formula for $\sin(A - B)$

$\sin(A - B) = \sin A \cos B - \cos A \sin B$

2. Substitute the given values and solve for $\cos A \sin B$

$\frac{3}{8} = \frac{5}{8} - \cos A \sin B$
$\cos A \sin B = \frac{5}{8} - \frac{3}{8} = \frac{2}{8} = \frac{1}{4}$

1. Express all terms using sine and cosine, and apply the double angle formula

$2(2\sin \theta \cos \theta)\left(\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}\right)$

2. Combine the terms in the bracket and simplify

$= 4\sin \theta \cos \theta \left(\frac{1 - \sin \theta}{\cos \theta}\right)$
$= 4\sin \theta (1 - \sin \theta)$

3. Expand to form a quadratic expression in $\sin \theta$

$= 4\sin \theta - 4\sin^2 \theta$

1. Substitute the quadratic expression from part (b) into the equation

From part (b), $2\sin 2\theta(\sec \theta - \tan \theta) = 4\sin\theta - 4\sin^2\theta$.
Substituting this into the equation:
$4\sin\theta - 4\sin^2\theta + 3 = 0$
$4\sin^2\theta - 4\sin\theta - 3 = 0$

2. Solve the quadratic equation for $\sin \theta$

$(2\sin\theta - 3)(2\sin\theta + 1) = 0$
$\sin\theta = \frac{3}{2} \quad \text{or} \quad \sin\theta = -\frac{1}{2}$
Since $-1 \le \sin\theta \le 1$, $\sin\theta = \frac{3}{2}$ has no solution.

3. Find the exact values of $\theta$ in the given range

For $\sin\theta = -\frac{1}{2}$, the basic angle is $\alpha = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}$.
Since $\sin\theta$ is negative, $\theta$ lies in the 3rd and 4th quadrants.
$\theta = \pi + \frac{\pi}{6} \quad \text{or} \quad \theta = 2\pi - \frac{\pi}{6}$
$\theta = \frac{7\pi}{6}, \frac{11\pi}{6}$