Show that the response function

$R(\theta )=\frac{1}{2}\sin 2\theta \sec \theta +2\cos 2\theta \csc \theta$

can be simplified to the form

$R(\theta )=2\csc \theta -3\sin \theta$

where $\sin \theta \ne 0$ and $\cos \theta \ne 0$.

The engineer needs to find the phase angles where the response is exactly 5. A student attempts to solve the equation

$\frac{1}{2}\sin 2\theta \sec \theta +2\cos 2\theta \csc \theta =5$

for $0^{\circ }\le \theta \le 360^{\circ }$. They use the result from part (a) to produce the following steps:

Step 1: $2\csc \theta -3\sin \theta =5$

Step 2: $\frac{2}{\sin \theta }-3\sin \theta =5$

Step 3: $3{\sin }^2\theta +5\sin \theta -2=0$

Step 4: $\sin \theta =\frac{1}{3}$ or $\sin \theta =-2$

Step 5: $\theta =19.5^{\circ },160.5^{\circ }$ (to 1 decimal place)

Explain why the value $\sin \theta =-2$ must be rejected in Step 4.

Determine if there are any further reasons, based on the domain constraints of the original filter model, why specific solutions might need to be rejected, and state the final correct solutions for $0^{\circ }\le \theta \le 360^{\circ }$.