Answer: (a) \(\displaystyle y=\frac{2}{\sqrt{4x-x^2-3}}\); (b) \(\displaystyle \phi=-\frac{10\pi}{3},-\frac{4\pi}{3},\frac{2\pi}{3},\frac{8\pi}{3}\).

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) From

\(2+\cos\theta=x,\)

we get

\(\cos\theta=x-2.\)

From

\(2\operatorname{cosec}\theta=y,\)

we have

\(\operatorname{cosec}\theta=\frac{y}{2},\)

so

\(\sin\theta=\frac{2}{y}.\)

Use the identity

\(\sin^2\theta+\cos^2\theta=1.\)

This gives

\(\left(\frac{2}{y}\right)^2+(x-2)^2=1.\)

So

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

Therefore

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

Since \(y\gt 2\), take the positive square root:

\(y=\frac{2}{\sqrt{1-(x-2)^2}}.\)

Equivalently,

\(y=\frac{2}{\sqrt{4x-x^2-3}}.\)

(b) Let

\(u=\frac{\phi}{2}.\)

The equation becomes

\(3\cos u=\sqrt3\sin u.\)

So

\(\tan u=\frac{3}{\sqrt3}=\sqrt3.\)

Therefore

\(u=\frac{\pi}{3}+n\pi,\)

where \(n\) is an integer. Hence

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

Now require

\(-4\pi\lt \phi\lt 4\pi.\)

The values that satisfy this are