(i) Start with the equation:

\(\sin(x - 60^\circ) - \cos(30^\circ - x) = 1\)

Use the identity \(\sin(a - b) = \sin a \cos b - \cos a \sin b\):

\(\sin(x - 60^\circ) = \sin x \cos 60^\circ - \cos x \sin 60^\circ\)

\(\cos(30^\circ - x) = \cos 30^\circ \cos x + \sin 30^\circ \sin x\)

Substitute these into the equation:

\((\sin x \cos 60^\circ - \cos x \sin 60^\circ) - (\cos 30^\circ \cos x + \sin 30^\circ \sin x) = 1\)

Using \(\cos 60^\circ = \frac{1}{2}\), \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\sin 30^\circ = \frac{1}{2}\):

\(\left(\sin x \cdot \frac{1}{2} - \cos x \cdot \frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x\right) = 1\)

Simplify:

\(\sin x \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} \cos x - \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x = 1\)

\(-\sqrt{3} \cos x = 1\)

\(\cos x = -\frac{1}{\sqrt{3}}\)

Thus, \(k = -\frac{1}{\sqrt{3}}\).

(ii) Solve \(\cos x = -\frac{1}{\sqrt{3}}\) for \(0^\circ < x < 180^\circ\):

The angle \(x\) where \(\cos x = -\frac{1}{\sqrt{3}}\) is approximately \(x = 125.3^\circ\).