(i) Differentiate \(x = t - \tan t\) with respect to \(t\):

\(\frac{dx}{dt} = 1 - \sec^2 t\).

Differentiate \(y = \ln(\cos t)\) with respect to \(t\):

\(\frac{dy}{dt} = \frac{-\sin t}{\cos t} = -\tan t\).

Using the chain rule, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\):

\(\frac{dy}{dx} = \frac{-\tan t}{1 - \sec^2 t} = \cot t\).

(ii) To find the \(x\)-coordinate where the gradient is 2, set \(\frac{dy}{dx} = 2\):

\(\cot t = 2 \Rightarrow t = \arctan\left(\frac{1}{2}\right)\).

Substitute \(t = \arctan\left(\frac{1}{2}\right)\) into \(x = t - \tan t\):

\(x = \arctan\left(\frac{1}{2}\right) - \frac{1}{2}\).

Calculate \(x \approx -0.0364\) to 3 significant figures.