[Cambridge/9709/2007]
The equation of a curve is $y = 2x - \tan x$, where $x$ is in radians. Find the coordinates of the stationary points of the curve for which $-\frac{1}{2}\pi < x < \frac{1}{2}\pi$.

1. Find the first derivative $\frac{dy}{dx}$ of the given curve equation.

$\frac{dy}{dx} = 2 - \sec^2 x$

2. Set the derivative to zero to locate the stationary points and simplify.

For stationary points, $\frac{dy}{dx} = 0$:
$2 - \sec^2 x = 0$
$\sec^2 x = 2$
$\cos^2 x = \frac{1}{2}$

3. Solve the trigonometric equation for $x$ within the specified domain.

$\cos x = \pm \frac{1}{\sqrt{2}}$
Since the domain is $-\frac{1}{2}\pi < x < \frac{1}{2}\pi$, we know that $\cos x > 0$.
Therefore, $\cos x = \frac{1}{\sqrt{2}}$.
$x = \frac{\pi}{4} \text{ or } x = -\frac{\pi}{4}$

4. Substitute the $x$-values into the original curve equation to find the corresponding $y$-coordinates.

For $x = \frac{\pi}{4}$:
$y = 2\left(\frac{\pi}{4}\right) - \tan\left(\frac{\pi}{4}\right) = \frac{\pi}{2} - 1$

For $x = -\frac{\pi}{4}$:
$y = 2\left(-\frac{\pi}{4}\right) - \tan\left(-\frac{\pi}{4}\right) = -\frac{\pi}{2} - (-1) = 1 - \frac{\pi}{2}$