Prove that $\dfrac{\sec^2 x}{1 + \sec^2 x} = \dfrac{2}{\cos 2x + 3}$. Hence, solve the equation $\dfrac{\sec^2 2x}{1 + \sec^2 2x} = \dfrac{4}{7}$ for $0 \le x \le 3$. State the number of solutions for $-2\pi \le x \le 2\pi$.

1. Express the left-hand side in terms of cosine.

$\text{LHS} = \frac{\sec^2 x}{1 + \sec^2 x} = \frac{\frac{1}{\cos^2 x}}{1 + \frac{1}{\cos^2 x}}$

2. Simplify the complex fraction.

$\text{LHS} = \frac{\frac{1}{\cos^2 x}}{\frac{\cos^2 x + 1}{\cos^2 x}} = \frac{1}{\cos^2 x + 1}$

3. Use the double angle formula for cosine to express the denominator in terms of $\cos 2x$.

Since $\cos 2x = 2\cos^2 x - 1$, we have $\cos^2 x = \frac{\cos 2x + 1}{2}$.

Substitute this into the expression:
$\text{LHS} = \frac{1}{\frac{\cos 2x + 1}{2} + 1} = \frac{1}{\frac{\cos 2x + 1 + 2}{2}} = \frac{2}{\cos 2x + 3} = \text{RHS}$

1. Apply the identity from part (a) to the given equation.

From (a), $\dfrac{\sec^2 x}{1 + \sec^2 x} = \dfrac{2}{\cos 2x + 3}$.
Replacing $x$ with $2x$, we get:
$\dfrac{\sec^2 2x}{1 + \sec^2 2x} = \dfrac{2}{\cos 4x + 3}$
So the equation becomes:
$\dfrac{2}{\cos 4x + 3} = \dfrac{4}{7}$

2. Solve for $\cos 4x$.

$14 = 4(\cos 4x + 3)$
$14 = 4\cos 4x + 12$
$4\cos 4x = 2$
$\cos 4x = \frac{1}{2}$

3. Determine the basic angle and the range for $4x$.

Basic angle $\alpha = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.
Given $0 \le x \le 3$, the range for $4x$ is $0 \le 4x \le 12$.

4. Find all possible values for $4x$ and then $x$.

Since $\cos 4x$ is positive, $4x$ is in the 1st and 4th quadrants.
$4x = \frac{\pi}{3}, 2\pi - \frac{\pi}{3}, 2\pi + \frac{\pi}{3}, 4\pi - \frac{\pi}{3}$
$4x = \frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}$
(Note: $\frac{11\pi}{3} \approx 11.519 \le 12$, but the next value $4\pi + \frac{\pi}{3} = \frac{13\pi}{3} \approx 13.61 > 12$)
$x = \frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}, \frac{11\pi}{12}$
$x \approx 0.262, 1.31, 1.83, 2.88$ (to 3 sig. fig.)

1. Recall the simplified equation from part (b) and adjust the domain for the new variable.

From part (b), the equation simplifies to $\cos 4x = 0.5$.
The given domain is $-2\pi \le x \le 2\pi$.
Multiplying by 4, the domain for $4x$ is $-8\pi \le 4x \le 8\pi$.

2. Determine the number of solutions in the adjusted domain.

The interval $[-8\pi, 8\pi]$ has a total length of $16\pi$, which corresponds to $8$ complete cycles of $2\pi$.
In each $2\pi$ cycle, the equation $\cos \theta = 0.5$ has exactly 2 solutions.
Total number of solutions = $8 \times 2 = 16$.