Answer: (a) \(y=(x-4)^2-3\); (b) \(\phi=-\frac{5\pi}{24},-\frac{\pi}{24},\frac{19\pi}{24},\frac{23\pi}{24}\).

Reduce the equation to a standard trigonometric form, find the principal solutions, then add the correct multiples of the period to cover the required interval.

(a) Since

\(\operatorname{sec}\theta=x-4,\)

we have

\(\tan^2\theta=\operatorname{sec}^2\theta-1=(x-4)^2-1.\)

Therefore

\(\operatorname{cot}^2\theta=\frac{1}{(x-4)^2-1}.\)

But \(\operatorname{cot}^2\theta=\frac1{y+2}\), so

\(\frac1{y+2}=\frac1{(x-4)^2-1}.\)

Hence

\(y+2=(x-4)^2-1,\)

and so

\(y=(x-4)^2-3.\)

(b) The equation is

\(\sqrt3\,\operatorname{cosec}\left(2\phi+\frac{3\pi}{4}\right)=2.\)

So

\(\operatorname{cosec}\left(2\phi+\frac{3\pi}{4}\right)=\frac{2}{\sqrt3}.\)

Hence

\(\sin\left(2\phi+\frac{3\pi}{4}\right)=\frac{\sqrt3}{2}.\)

Therefore

\(2\phi+\frac{3\pi}{4}=\frac{\pi}{3}+2k\pi\)

or

\(2\phi+\frac{3\pi}{4}=\frac{2\pi}{3}+2k\pi.\)

This gives

\(\phi=-\frac{5\pi}{24}+k\pi\)

or

\(\phi=-\frac{\pi}{24}+k\pi.\)

For \(-\pi\lt\phi\lt\pi\), the solutions are

\(\phi=-\frac{5\pi}{24},\ -\frac{\pi}{24},\ \frac{19\pi}{24},\ \frac{23\pi}{24}.\)