Answer: \(x=78.5^\circ,\ 120^\circ,\ 240^\circ,\ 281.5^\circ\).

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:

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

Use

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

Then

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

Also,

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

Since \(\cos x-\sin x=-(\sin x-\cos x)\),

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

Therefore

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

Now

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

So

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

(b) Write the equation in terms of \(\sin x\) and \(\cos x\):

\(9\frac{\cos x}{\sin x}+3\frac1{\sin x}=\frac{\sin x}{\cos x}.\)

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

\(9\cos^2x+3\cos x=\sin^2x.\)

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

\(9\cos^2x+3\cos x=1-\cos^2x.\)

So

\(10\cos^2x+3\cos x-1=0.\)

Factorise:

\((5\cos x-1)(2\cos x+1)=0.\)

Therefore

\(\cos x=\frac15\quad\text{or}\quad \cos x=-\frac12.\)

For \(0^\circ\lt x\lt360^\circ\), these give

\(x=78.5^\circ,\ 281.5^\circ,\ 120^\circ,\ 240^\circ.\)

So the solutions are

\(x=78.5^\circ,\ 120^\circ,\ 240^\circ,\ 281.5^\circ.\)

This completes the solution and gives the required result.