Answer: (a) \(\theta=-161.6^\circ,-63.4^\circ,18.4^\circ,116.6^\circ\). (b) \(\phi=0.456,2.76,3.60\).

Use the standard trigonometric identities and make sure the final angles are chosen from the interval given in the question.

For part (a), use \(\csc^2\theta=\cot^2\theta+1\).

\(2(\cot^2\theta+1)-5=5\cot\theta\)

So

\(2\cot^2\theta-5\cot\theta-3=0\)

Factorising,

\((2\cot\theta+1)(\cot\theta-3)=0\)

Thus \(\cot\theta=-\frac12\) or \(\cot\theta=3\), so

\(\tan\theta=-2\) or \(\tan\theta=\frac13\).

In the interval \(-180^\circ\leq\theta\leq180^\circ\), the solutions are

\(\theta=-161.6^\circ,\,-63.4^\circ,\,18.4^\circ,\,116.6^\circ\).

For part (b),

\(3\sin(2\phi+1.5)=2\)

so

\(\sin(2\phi+1.5)=\frac23\).

Let \(u=2\phi+1.5\). Since \(0<\phi<5\),

\(1.5

The values of \(u\) in this interval are approximately

\(2.4119,\quad 7.0129,\quad 8.6951\).

Since \(\phi=\frac{u-1.5}{2}\),

we get

\(\phi=0.456,\quad 2.76,\quad 3.60\)

to 3 significant figures.

This gives the required answer:

(a) \(\theta=-161.6^\circ,-63.4^\circ,18.4^\circ,116.6^\circ\).

(b) \(\phi=0.456,2.76,3.60\).