Answer: (a) shown; (b) \(x=90^\circ,\ 221.8^\circ,\ 318.2^\circ\).

Answer: (a) shown; (b) \(x=90^\circ,\ 221.8^\circ,\ 318.2^\circ\).

(a) Start with the left-hand side:

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

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

\(\frac{\sin^{2}x+(1-\cos x)^2}{(1-\cos x)\sin x}.\)

Expand the numerator:

\(\sin^{2}x+1-2\cos x+ \cos^{2}x.\)

Since \(\sin^{2}x+ \cos^{2}x=1\), this becomes

\(2-2\cos x=2(1-\cos x).\)

Therefore

Cancel \((1-\cos x)\):

\(=\frac2{\sin x}=2\operatorname{cosec}x.\)

(b) By part (a), the equation becomes

\(2\operatorname{cosec}x=3\sin x-1.\)

Since \(\operatorname{cosec}x=\frac1{\sin x}\),

\(\frac2{\sin x}=3\sin x-1.\)

Multiply by \(\sin x\):

\(2=3\sin^{2}x-\sin x.\)

So

\(3\sin^{2}x-\sin x-2=0.\)

Factorising,

\((3\sin x+2)(\sin x-1)=0.\)

Hence

\(\sin x=1 \quad\text{or}\quad \sin x=-\frac23.\)

For \(0^\circ\lt x\lt 360^\circ\), this gives

\(x=90^\circ, \quad x=221.8^\circ, \quad x=318.2^\circ.\)