Answer: (a) \(x=-110^\circ,-50^\circ,10^\circ,70^\circ\). (b) \(y=0,\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\).

(a) Start with

\(3\sec3x=\sqrt3\operatorname{cosec}3x.\)

Using \(\sec3x=\frac{1}{\cos3x}\) and \(\operatorname{cosec}3x=\frac{1}{\sin3x}\),

\(\frac{3}{\cos3x}=\frac{\sqrt3}{\sin3x}.\)

So

\(3\sin3x=\sqrt3\cos3x,\)

and hence

\(\tan3x=\frac{1}{\sqrt3}.\)

Since \(-120^\circ\leqslant x\leqslant120^\circ\),

\(-360^\circ\leqslant3x\leqslant360^\circ.\)

The valid values are

\(3x=-330^\circ,-150^\circ,30^\circ,210^\circ.\)

Therefore

\(x=-110^\circ,-50^\circ,10^\circ,70^\circ.\)

(b) Let \(u=y+\frac{\pi}{3}\). The equation becomes

\(2\cos u\sin u=\sin u.\)

So

\(\sin u(2\cos u-1)=0.\)

Hence either \(\sin u=0\), or \(\cos u=\frac12\).

Since \(0\leqslant y\lt 2\pi\),

\(\frac{\pi}{3}\leqslant u\lt \frac{7\pi}{3}.\)

The valid values of \(u\) are

\(\frac{\pi}{3},\pi,\frac{5\pi}{3},2\pi.\)

Subtracting \(\frac{\pi}{3}\),

\(y=0,\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}.\)