(i) Expand \(\cos(x + 45^\circ)\):

\(\cos(x + 45^\circ) = \frac{\cos x}{\sqrt{2}} - \frac{\sin x}{\sqrt{2}}\)

Express \(\cos(x + 45^\circ) - \sqrt{2} \sin x\):

\(\frac{\cos x}{\sqrt{2}} - \frac{\sin x}{\sqrt{2}} - \sqrt{2} \sin x = \frac{\cos x}{\sqrt{2}} - \left(\frac{1}{\sqrt{2}} + \sqrt{2}\right) \sin x\)

Let \(a = \frac{1}{\sqrt{2}}\) and \(b = -\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)\). Then:

\(R = \sqrt{a^2 + b^2} = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)\right)^2} = 2.236\)

\(\tan \alpha = \frac{b}{a} = \frac{-\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)}{\frac{1}{\sqrt{2}}}\)

\(\alpha = \tan^{-1}\left(\frac{-\left(\frac{1}{\sqrt{2}} + \sqrt{2}\right)}{\frac{1}{\sqrt{2}}}\right) = 71.57^\circ\)

(ii) Solve \(\cos(x + 45^\circ) - \sqrt{2} \sin x = 2\):

Using the expression from (i):

\(R \cos(x + \alpha) = 2\)

\(2.236 \cos(x + 71.57^\circ) = 2\)

\(\cos(x + 71.57^\circ) = \frac{2}{2.236}\)

\(x + 71.57^\circ = \cos^{-1}\left(\frac{2}{2.236}\right)\)

\(x + 71.57^\circ = 26.57^\circ \text{ or } 333.43^\circ\)

\(x = 315^\circ \text{ or } 261.9^\circ\)