Prove the identity $\frac{\cos x}{\text{cosec } x - 1} + \frac{\cos x}{\text{cosec } x + 1} = 2 \tan x$. Solve the equation $\tan x + 2\sec^2 x = 5$ for $0 \le x \le 2\pi$.

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

$\text{LHS} = \frac{\cos x(\text{cosec } x + 1) + \cos x(\text{cosec } x - 1)}{(\text{cosec } x - 1)(\text{cosec } x + 1)}$

2. Simplify the numerator and expand the denominator.

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

3. Apply the Pythagorean identity to the denominator.

Since $\text{cosec}^2 x - 1 = \cot^2 x$,
$\text{LHS} = \frac{2\cos x \text{cosec } x}{\cot^2 x}$

4. Express in terms of sine and cosine to simplify to the right-hand side.

$\text{LHS} = \frac{2\cos x \left(\frac{1}{\sin x}\right)}{\frac{\cos^2 x}{\sin^2 x}}$
$= \frac{2\cos x}{\sin x} \times \frac{\sin^2 x}{\cos^2 x}$
$= \frac{2\sin x}{\cos x}$
$= 2\tan x = \text{RHS}$

1. Express the equation in terms of a single trigonometric function

Using the identity $\sec^2 x = 1 + \tan^2 x$:
$\tan x + 2(1 + \tan^2 x) = 5$
$2\tan^2 x + \tan x - 3 = 0$

2. Solve the quadratic equation for $\tan x$

$(2\tan x + 3)(\tan x - 1) = 0$
$\tan x = -1.5 \quad \text{or} \quad \tan x = 1$

3. Find the reference angles

For $\tan x = 1$, the reference angle is $\frac{\pi}{4}$.
For $\tan x = -1.5$, the reference angle is $\tan^{-1}(1.5) \approx 0.98279$.

4. Find all solutions in the given interval $0 \le x \le 2\pi$

For $\tan x = 1$ (1st and 3rd quadrants):
$x = \frac{\pi}{4}, \pi + \frac{\pi}{4}$
$x \approx 0.7854, 3.9270$

For $\tan x = -1.5$ (2nd and 4th quadrants):
$x = \pi - 0.98279, 2\pi - 0.98279$
$x \approx 2.1588, 5.3004$

5. State the final answers rounded to 3 significant figures

$x = 0.785, 2.16, 3.93, 5.30$