Answer: \(\theta=131.8^\circ,228.2^\circ,210^\circ,330^\circ\); \(\phi=-\frac{5\pi}{24},-\frac{\pi}{24},\frac{19\pi}{24},\frac{23\pi}{24}\).

(a)(i) Factor by grouping:

\(6xy+3y+4x+2=3y(2x+1)+2(2x+1).\)

So

\(6xy+3y+4x+2=(3y+2)(2x+1).\)

(a)(ii) Use the factorisation with \(x=\sin\theta\) and \(y=\cos\theta\):

Therefore

\((3\cos\theta+2)(2\sin\theta+1)=0.\)

So

\(\cos\theta=-\frac23\)

or

\(\sin\theta=-\frac12.\)

For \(0^\circ\lt\theta\lt360^\circ\), \(\cos\theta=-\frac23\) gives

\(\theta=131.8^\circ,\ 228.2^\circ.\)

Also, \(\sin\theta=-\frac12\) gives

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

(b) The equation is

So

Hence

Let

\(u=2\phi+\frac{\pi}{4}.\)

Since \(-\pi\lt\phi\lt\pi\),

\(-\frac{7\pi}{4}\lt u\lt \frac{9\pi}{4}.\)

In this interval, \(\cos u=\frac{\sqrt3}{2}\) when

Now solve \(2\phi+\frac{\pi}{4}=u\). This gives