Answer: \(\sin x(\operatorname{cot} x+\tan x)=\operatorname{sec} x\), so the solutions are \(x=60^\circ,120^\circ,240^\circ,300^\circ\).

(i) Start from the left-hand side:

\(\sin x(\operatorname{cot} x+\tan x) =\sin x\left(\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\right).\)

Distribute \(\sin x\):

\(\sin x(\operatorname{cot} x+\tan x) =\cos x+\frac{\sin^2x}{\cos x}.\)

Write over a common denominator:

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

Using \(\cos^2x+\sin^2x=1\), this becomes

\(\frac{1}{\cos x}=\operatorname{sec} x.\)

Hence

\(\sin x(\operatorname{cot} x+\tan x)=\operatorname{sec} x.\)

(ii) Using part (i), the equation becomes

\(|\operatorname{sec} x|=2.\)

So

\(\operatorname{sec} x=2 \quad\text{or}\quad \operatorname{sec} x=-2.\)

Equivalently,

\(\cos x=\frac12 \quad\text{or}\quad \cos x=-\frac12.\)

For \(0^\circ\le x\le360^\circ\):

\(\cos x=\frac12 \Rightarrow x=60^\circ,300^\circ,\)

\(\cos x=-\frac12 \Rightarrow x=120^\circ,240^\circ.\)

Therefore the solutions are

\(\boxed{x=60^\circ,120^\circ,240^\circ,300^\circ}.\)