Prove the identity $\frac{1 + \sin x}{\cos x} + \frac{\cos x}{1 + \sin x} = 2 \sec x$. Hence, solve the equation $\frac{1 + \sin x}{\cos x} + \frac{\cos x}{1 + \sin x} = 4$ for $0 < x < 2\pi$.

1. Combine the fractions on the left-hand side using a common denominator.

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

2. Expand the numerator.

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

3. Apply the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to simplify the numerator.

$\text{LHS} = \frac{1 + 2\sin x + 1}{\cos x (1 + \sin x)} = \frac{2 + 2\sin x}{\cos x (1 + \sin x)}$

4. Factorise the numerator, cancel the common factor, and apply the reciprocal identity to reach the RHS.

$\text{LHS} = \frac{2(1 + \sin x)}{\cos x (1 + \sin x)} = \frac{2}{\cos x} = 2\sec x = \text{RHS}$

1. Substitute the identity proven in part (i) into the equation

From (i), $\frac{1 + \sin x}{\cos x} + \frac{\cos x}{1 + \sin x} = 2 \sec x$.

So, $2 \sec x = 4$

2. Simplify the equation to find $\cos x$

$\sec x = 2$

$\frac{1}{\cos x} = 2$

$\cos x = \frac{1}{2}$

3. Solve for $x$ in the given domain $0 < x < 2\pi$

Basic angle $\alpha = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$

Since $\cos x$ is positive, $x$ is in the 1st and 4th quadrants.

$x = \frac{\pi}{3}$ or $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$