Answer: \(x=60^\circ,\ 180^\circ,\ 300^\circ\).

We start with the main method. Rewrite everything in terms of \(\sin x\) and \(\cos x\), then simplify using standard identities.

(a) Start with the left-hand side:

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

Use \(\tan x=\frac{\sin x}{\cos x}\):

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

This becomes

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

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

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

The denominator is

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

The numerator is

\(\sin^2x\cos x+\sin x\cos^2x-\sin x\cos^2x+\cos^3x.\)

This simplifies to

\(\sin^2x\cos x+\cos^3x=\cos x(\sin^2x+\cos^2x)=\cos x.\)

Therefore

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

(b) The equation becomes

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

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

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

So

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

Hence

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

Rearranging gives

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

Factorise:

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

Thus

\(\cos x=\frac12\quad\text{or}\quad \cos x=-1.\)

For \(0^\circ\lt x\lt360^\circ\), this gives

\(x=60^\circ,\ 180^\circ,\ 300^\circ.\)

This completes the solution and gives the required result.