First, find \(\frac{dy}{d\theta}\):

\(\frac{dy}{d\theta} = 1 - 2\sin \theta\).

Next, find \(\frac{dx}{d\theta}\) using the quotient rule:

\(x = \frac{\cos \theta}{2 - \sin \theta}\).

Let \(u = \cos \theta\) and \(v = 2 - \sin \theta\).

Then \(\frac{du}{d\theta} = -\sin \theta\) and \(\frac{dv}{d\theta} = -\cos \theta\).

Using the quotient rule, \(\frac{dx}{d\theta} = \frac{v \frac{du}{d\theta} - u \frac{dv}{d\theta}}{v^2}\).

\(\frac{dx}{d\theta} = \frac{(2 - \sin \theta)(-\sin \theta) + \cos \theta \cdot \cos \theta}{(2 - \sin \theta)^2}\).

\(\frac{dx}{d\theta} = \frac{-2\sin \theta + \sin^2 \theta + \cos^2 \theta}{(2 - \sin \theta)^2}\).

Since \(\cos^2 \theta + \sin^2 \theta = 1\),

\(\frac{dx}{d\theta} = \frac{-2\sin \theta + 1}{(2 - \sin \theta)^2}\).

Now, use \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}\):

\(\frac{dy}{dx} = \frac{1 - 2\sin \theta}{\frac{-2\sin \theta + 1}{(2 - \sin \theta)^2}}\).

\(\frac{dy}{dx} = (2 - \sin \theta)^2\).