Prove the identity $\dfrac{\cos^{2} x - \sin^{2} x}{1 + 2\sin x\cos x} = \dfrac{1 - \tan x}{1 + \tan x}$. Solve the equation $\dfrac{\cos^{2} x - \sin^{2} x}{1 + 2\sin x\cos x} = \dfrac{2}{3}\tan x$ for $0 \le x \le 2\pi$.

1. Factorise the numerator using the difference of two squares

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

2. Rewrite the denominator using the Pythagorean identity

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

3. Factorise the denominator as a perfect square

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

4. Cancel the common factor and divide numerator and denominator by $\cos x$ to obtain $\tan x$

$\text{LHS} = \frac{\cos x - \sin x}{\cos x + \sin x} = \frac{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}} = \frac{1 - \tan x}{1 + \tan x} = \text{RHS}$

1. Use the identity from part (a) to rewrite the equation in terms of $\tan x$

Using the result from (a),
$\frac{1 - \tan x}{1 + \tan x} = \frac{2}{3}\tan x$

2. Form a quadratic equation in $\tan x$

$3(1 - \tan x) = 2\tan x(1 + \tan x)$
$3 - 3\tan x = 2\tan x + 2\tan^2 x$
$2\tan^2 x + 5\tan x - 3 = 0$

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

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

4. Solve for $x$ in the given domain $0 \le x \le 2\pi$

For $\tan x = 0.5$:
Basic angle $\alpha = \tan^{-1}(0.5) \approx 0.4636$
$x$ is in the 1st and 3rd quadrants:
$x = 0.4636, \pi + 0.4636$
$x \approx 0.464, 3.61$

For $\tan x = -3$:
Basic angle $\beta = \tan^{-1}(3) \approx 1.2490$
$x$ is in the 2nd and 4th quadrants:
$x = \pi - 1.2490, 2\pi - 1.2490$
$x \approx 1.89, 5.03$

$x = 0.464, 1.89, 3.61, 5.03$