Answer: \(x=48.6^\circ,131.4^\circ,210^\circ,330^\circ\).

(a) Start with the left-hand side:

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

Since

\(\tan x=\frac{\sin x}{\cos x},\)

we have

\(\frac{\sin x\tan x}{1-\cos x} =\frac{\sin x\cdot\frac{\sin x}{\cos x}}{1-\cos x} =\frac{\sin^2x}{\cos x(1-\cos x)}.\)

Use

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

Then

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

Factorise the numerator:

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

Therefore

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

So

\(\frac{1+\cos x}{\cos x}=1+\frac1{\cos x}=1+\operatorname{sec} x.\)

This proves the identity.

(b) Rewrite the equation using sine and cosine:

\(5\frac{\sin x}{\cos x}-3\frac{\cos x}{\sin x} =2\frac1{\cos x}.\)

Multiply by \(\sin x\cos x\):

\(5\sin^2x-3\cos^2x=2\sin x.\)

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

\(5\sin^2x-3(1-\sin^2x)=2\sin x.\)

So

\(8\sin^2x-2\sin x-3=0.\)

Factorise:

\((2\sin x+1)(4\sin x-3)=0.\)

Thus

\(\sin x=-\frac12\)

or

\(\sin x=\frac34.\)

For \(0^\circ\leq x\leq360^\circ\),

\(\sin x=-\frac12\)

gives

\(x=210^\circ,\ 330^\circ.\)

Also,

\(\sin x=\frac34\)

gives

\(x=48.6^\circ,\ 131.4^\circ.\)

Therefore the solutions are

\(x=48.6^\circ,\ 131.4^\circ,\ 210^\circ,\ 330^\circ.\)