(i) Start with the equation \(2 \sin x \tan x + 3 = 0\).

Using \(\tan x = \frac{\sin x}{\cos x}\), rewrite the equation as:

\(2 \sin x \left(\frac{\sin x}{\cos x}\right) + 3 = 0\)

\(\frac{2 \sin^2 x}{\cos x} + 3 = 0\)

Using the identity \(\sin^2 x = 1 - \cos^2 x\), substitute:

\(\frac{2(1 - \cos^2 x)}{\cos x} + 3 = 0\)

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

\(2(1 - \cos^2 x) + 3 \cos x = 0\)

\(2 - 2 \cos^2 x + 3 \cos x = 0\)

Rearrange to get:

\(2 \cos^2 x - 3 \cos x - 2 = 0\)

(ii) Solve the quadratic equation \(2 \cos^2 x - 3 \cos x - 2 = 0\).

Factor the quadratic:

\((2 \cos x + 1)(\cos x - 2) = 0\)

Set each factor to zero:

\(2 \cos x + 1 = 0\) gives \(\cos x = -\frac{1}{2}\)

\(\cos x - 2 = 0\) gives \(\cos x = 2\) (not possible since \(|\cos x| \leq 1\))

For \(\cos x = -\frac{1}{2}\), the solutions in the interval \(0^\circ \leq x \leq 360^\circ\) are:

\(x = 120^\circ\) and \(x = 240^\circ\).