Find all the angles between \(0^\circ\) and \(180^\circ\) which satisfy the equations

- \(\tan (x + 60^\circ) = 1\),

Let’s find the general solution. We know that \(\arctan{1} = 45 ^\circ\), so \[x+60=45+180n \implies x = -15 + 180n.\]

Recall that the function \(\tan x\) is periodic with period \(180^\circ\).

Thus the only solution in our range is \(165^\circ\).

- \(8 \sin y + 3 \sec y = 0\),

Note \(\cos y \neq 0\), since then \(\sec y\) would be undefined.

Now \(2\sin y\cos y = \sin(2y)\), and so on rearranging, our equation becomes \[ \sin(2y) = -\frac{3}{4}. \]

So the general solution here is \(2y = -48.6 + 360n\), or \(228.6+360n\).

This is since \(228.6 = 180 - (-48.6)\).

This implies \(y = -24.3 + 180n\), or \(114.3 + 180n\), and so the solutions in our range are \(114.3^\circ\) and \(155.7^\circ\).

- \(3 \sin^2 z - 8 \sin z \cos z - 3 \cos^2 z = 0\).

Could \(\cos z\) be \(0\) here? This would mean that for the equation to hold, \(\sin z = 0\). But \(\cos z\) and \(\sin z\) are never zero together, so \(\cos z\) cannot be zero if the equation holds.

So we can divide the equation by \(\cos^2 z\), giving us \(3 \tan^2 z - 8 \tan z - 3 = 0\), which becomes \((3\tan z + 1)(\tan z -3) = 0\).

So \(\tan z = -\dfrac{1}{3}\) or \(\tan z = 3\), and \(z = -18.4 + 180n\) or \(71.6 + 180n\).

Thus the two solutions in our range are \(71.6^\circ\) and \(161.6^\circ\).

Alternatively, we could factorise the original expression, \[3 \sin^2 z - 8 \sin z \cos z - 3 \cos^2 z =(3\sin z+\cos z)(\sin z-3\cos z)\] with the same result.