(i) Differentiate \(x = t + \cos t\) with respect to \(t\):

\(\frac{dx}{dt} = 1 - \sin t\).

Differentiate \(y = \ln(1 + \sin t)\) with respect to \(t\):

\(\frac{dy}{dt} = \frac{\cos t}{1 + \sin t}\).

Use the chain rule to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{\frac{\cos t}{1 + \sin t}}{1 - \sin t} = \frac{\cos t}{(1 + \sin t)(1 - \sin t)} = \frac{\cos t}{1 - \sin^2 t} = \frac{\cos t}{\cos^2 t} = \sec t\).

(ii) Set \(\frac{dy}{dx} = 3\):

\(\sec t = 3 \Rightarrow \cos t = \frac{1}{3}\).

Find \(t\) using \(t = \cos^{-1}\left(\frac{1}{3}\right)\).

Calculate \(x = t + \cos t\) for each \(t\):

For \(t = \cos^{-1}\left(\frac{1}{3}\right)\), \(x = 1.56\).

For \(t = -\cos^{-1}\left(\frac{1}{3}\right)\), \(x = -0.898\).