Answer: (a) \(y=\frac{7\pi}{12}\) or \(y=\frac{11\pi}{12}\). (b) \(z=80.4^\circ\) or \(z=279.6^\circ\).
(a) From
\(3\operatorname{cot}^2\left(y-\frac{\pi}{4}\right)=1,\)
we get
\(\operatorname{cot}^2\left(y-\frac{\pi}{4}\right)=\frac13.\)
So
\(\tan^2\left(y-\frac{\pi}{4}\right)=3.\)
Hence
\(\tan\left(y-\frac{\pi}{4}\right)=\pm\sqrt3.\)
Since \(0\lt y\lt\pi\),
\(-\frac{\pi}{4}\lt y-\frac{\pi}{4}\lt\frac{3\pi}{4}.\)
The valid angles are
\(y-\frac{\pi}{4}=\frac{\pi}{3}\quad\text{or}\quad \frac{2\pi}{3}.\)
Therefore
\(\boxed{y=\frac{7\pi}{12}\quad\text{or}\quad y=\frac{11\pi}{12}}.\)
(b) Write the equation in terms of sine and cosine:
\(7\frac{\cos z}{\sin z}+\frac{\sin z}{\cos z}=\frac7{\sin z}.\)
Multiply by \(\sin z\cos z\):
\(7\cos^2z+\sin^2z=7\cos z.\)
Using \(\sin^2z=1-\cos^2z\),
\(6\cos^2z-7\cos z+1=0.\)
Factorising,
\((6\cos z-1)(\cos z-1)=0.\)
The value \(\cos z=1\) would make the original equation not defined, so it is rejected. Thus
\(\cos z=\frac16.\)
For \(0^\circ\leq z\leq360^\circ\),
\(\boxed{z=80.4^\circ\quad\text{or}\quad z=279.6^\circ}.\)
