(i) Start with the equation \(1 + \sin x \tan x = 5 \cos x\).

Replace \(\tan x\) with \(\frac{\sin x}{\cos x}\):

\(1 + \frac{\sin^2 x}{\cos x} = 5 \cos x\)

Multiply through by \(\cos x\) to eliminate the fraction:

\(\cos x + \sin^2 x = 5 \cos^2 x\)

Use the identity \(\sin^2 x = 1 - \cos^2 x\):

\(\cos x + 1 - \cos^2 x = 5 \cos^2 x\)

Rearrange to form a quadratic equation:

\(6 \cos^2 x - \cos x - 1 = 0\)

(ii) Solve the quadratic equation \(6 \cos^2 x - \cos x - 1 = 0\).

Using the quadratic formula \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6\), \(b = -1\), \(c = -1\):

\(\cos x = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 6 \times (-1)}}{2 \times 6}\)

\(\cos x = \frac{1 \pm \sqrt{1 + 24}}{12}\)

\(\cos x = \frac{1 \pm 5}{12}\)

\(\cos x = \frac{6}{12} = \frac{1}{2}\) or \(\cos x = \frac{-4}{12} = -\frac{1}{3}\)

For \(\cos x = \frac{1}{2}\), \(x = 60^\circ\).

For \(\cos x = -\frac{1}{3}\), \(x \approx 109.5^\circ\).