Answer: \(\theta=-150^\circ,-30^\circ,30^\circ,150^\circ\); \(\phi=\dfrac{7\pi}{36},\dfrac{19\pi}{36}\).

(a)(i) Use \(\operatorname{cosec}\theta=\frac1{\sin\theta}\). Then

\((1+\operatorname{cosec}\theta)(\sin\theta-\sin^2\theta) =\left(1+\frac1{\sin\theta}\right)\sin\theta(1-\sin\theta).\)

This becomes

\(\frac{\sin\theta+1}{\sin\theta}\cdot \sin\theta(1-\sin\theta) =(1+\sin\theta)(1-\sin\theta).\)

So

\((1+\operatorname{cosec}\theta)(\sin\theta-\sin^2\theta) =1-\sin^2\theta =\cos^2\theta.\)

Taking the reciprocal gives

\(\frac{1}{(1+\operatorname{cosec}\theta)(\sin\theta-\sin^2\theta)} =\frac1{\cos^2\theta} =\operatorname{sec}^2\theta.\)

(a)(ii) From part (a)(i), if

\((1+\operatorname{cosec}\theta)(\sin\theta-\sin^2\theta)=\frac34,\)

then

\(\operatorname{sec}^2\theta=\frac{4}{3}.\)

Hence

\(\cos^2\theta=\frac34,\)

so

\(\cos\theta=\pm\frac{\sqrt3}{2}.\)

For \(-180^\circ\leq\theta\leq180^\circ\), this gives

\(\theta=-150^\circ,\ -30^\circ,\ 30^\circ,\ 150^\circ.\)

(b) Let

\(u=3\phi+\frac{2\pi}{3}.\)

The equation is \(\sin u=\cos u\), so

\(\tan u=1.\)

Since \(0\leq\phi\leq\frac{2\pi}{3}\),

\(\frac{2\pi}{3}\leq u\leq\frac{8\pi}{3}.\)

In this interval,

\(u=\frac{5\pi}{4}\quad\text{or}\quad u=\frac{9\pi}{4}.\)

Therefore

\(3\phi=\frac{5\pi}{4}-\frac{2\pi}{3}=\frac{7\pi}{12},\)

or

\(3\phi=\frac{9\pi}{4}-\frac{2\pi}{3}=\frac{19\pi}{12}.\)

Hence

\(\phi=\frac{7\pi}{36}\quad\text{or}\quad \phi=\frac{19\pi}{36}.\)