Answer: \(\theta=96.4^\circ\) or \(\theta=623.6^\circ\).

(a) Start with the left-hand side:

\(\dfrac{1-\sin x}{\cos x}+\dfrac{\cos x}{1-\sin x}.\)

Use the common denominator \(\cos x(1-\sin x)\):

\(\dfrac{(1-\sin x)^2+\cos^2x}{\cos x(1-\sin x)}.\)

Now expand the numerator:

\((1-\sin x)^2+\cos^2x=1-2\sin x+\sin^2x+\cos^2x.\)

Since \(\sin^2x+\cos^2x=1\), this becomes

\(2-2\sin x=2(1-\sin x).\)

Therefore

\(\dfrac{(1-\sin x)^2+\cos^2x}{\cos x(1-\sin x)}=\dfrac{2(1-\sin x)}{\cos x(1-\sin x)}=\dfrac2{\cos x}=2\sec x.\)

This proves the identity.

(b) Using the identity with \(x=\dfrac\theta2\), the equation becomes

\(2\sec\dfrac\theta2=3.\)

Hence

\(\sec\dfrac\theta2=\dfrac32,\)

so

\(\cos\dfrac\theta2=\dfrac23.\)

Since

\(0^\circ\leqslant\theta\leqslant720^\circ,\)

we have

\(0^\circ\leqslant\dfrac\theta2\leqslant360^\circ.\)

In this interval, \(\cos u=\dfrac23\) has two solutions:

\(u=48.189\ldots^\circ\quad\text{or}\quad u=311.810\ldots^\circ.\)

Therefore

\(\theta=2u=96.378\ldots^\circ\quad\text{or}\quad623.621\ldots^\circ.\)

To one decimal place,

\(\theta=96.4^\circ\quad\text{or}\quad\theta=623.6^\circ.\)