Answer: (a) \(x=\dfrac{\pi}{12},\ \dfrac{5\pi}{12}\). (b) \(y=53.1^\circ,\ 233.1^\circ\). (c) \(z=60^\circ,\ 104.5^\circ,\ 255.5^\circ,\ 300^\circ\).

(a) Divide by \(2\):

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

For \(0\lt x\lt \pi\),

\(\frac{\pi}{4}\lt x+\frac{\pi}{4}\lt \frac{5\pi}{4}.\)

In this interval,

\(x+\frac{\pi}{4}=\frac{\pi}{3} \quad\text{or}\quad x+\frac{\pi}{4}=\frac{2\pi}{3}.\)

Therefore

\(x=\frac{\pi}{12} \quad\text{or}\quad x=\frac{5\pi}{12}.\)

(b) Use

\(\operatorname{sec} y=\frac1{\cos y}, \qquad \operatorname{cosec}y=\frac1{\sin y}.\)

The equation becomes

\(\frac{3}{\cos y}=\frac{4}{\sin y}.\)

So

\(3\sin y=4\cos y.\)

Hence

\(\tan y=\frac43.\)

For \(0^\circ\lt y\lt 360^\circ\), this gives

\(y=53.1^\circ,\ 233.1^\circ.\)

(c) Write the equation in sine and cosine:

\(7\frac{\cos z}{\sin z}-\frac{\sin z}{\cos z} = \frac{2}{\sin z}.\)

Multiply by \(\sin z\cos z\):

\(7\cos^2z-\sin^2z=2\cos z.\)

Use \(\sin^2z=1-\cos^2z\):

\(7\cos^2z-(1-\cos^2z)=2\cos z.\)

So

\(8\cos^2z-2\cos z-1=0.\)

Factorise:

\((4\cos z+1)(2\cos z-1)=0.\)

Thus

\(\cos z=\frac12 \quad\text{or}\quad \cos z=-\frac14.\)

For \(0^\circ\lt z\lt 360^\circ\),

\(z=60^\circ,\ 300^\circ\)

from \(\cos z=\dfrac12\), and

\(z=104.5^\circ,\ 255.5^\circ\)

from \(\cos z=-\dfrac14\).