Answer: \(x=0.955,\ 2.19,\ 4.10,\ 5.33\) radians.

(i) Start with the left-hand side:

\(\frac{1}{1-\cos x}-\frac{1}{1+\cos x}.\)

Use the common denominator \((1-\cos x)(1+\cos x)\):

\(\frac{1+\cos x-(1-\cos x)}{(1-\cos x)(1+\cos x)}.\)

The numerator is \(2\cos x\), and the denominator is \(1-\cos^2x=\sin^2x\). Hence

\(\frac{2\cos x}{\sin^2x} =2\left(\frac1{\sin x}\right)\left(\frac{\cos x}{\sin x}\right) =2\operatorname{cosec}x\operatorname{cot} x.\)

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

\(2\operatorname{cosec}x\operatorname{cot} x=\operatorname{sec} x.\)

Write this in terms of \(\sin x\) and \(\cos x\):

\(2\frac{1}{\sin x}\frac{\cos x}{\sin x} = \frac1{\cos x}.\)

Therefore

\(\frac{2\cos x}{\sin^2x}=\frac1{\cos x}.\)

Multiplying by \(\sin^2x\cos x\) gives

\(2\cos^2x=\sin^2x.\)

Since \(\sin^2x=1-\cos^2x\),

\(2\cos^2x=1-\cos^2x, \qquad 3\cos^2x=1.\)

So

\(\cos x=\pm\frac1{\sqrt3}.\)

For \(0\leq x\leq2\pi\), this gives

\(\boxed{x=0.955,\ 2.19,\ 4.10,\ 5.33}.\)