(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 \(x = 125.3^\circ\).