Answer: \(\theta=-\dfrac{7\pi}{18},\ \dfrac{\pi}{18},\ \dfrac{5\pi}{18}\).
We start with the main method. Factorise the trigonometric equation first, then solve each factor over the required interval.
Start with
\(\sec\left(3\theta-\frac{\pi}{2}\right)=2.\)
Since \(\sec u=\frac1{\cos u}\), this is equivalent to
\(\cos\left(3\theta-\frac{\pi}{2}\right)=\frac12.\)
Let
\(u=3\theta-\frac{\pi}{2}.\)
The given range is
\(-\frac{\pi}{2}\leqslant\theta\leqslant\frac{\pi}{2}.\)
Therefore
\(-2\pi\leqslant u\leqslant\pi.\)
In this interval, \(\cos u=\frac12\) when
\(u=-\frac{5\pi}{3},\quad -\frac{\pi}{3},\quad \frac{\pi}{3}.\)
Now solve \(3\theta-\frac{\pi}{2}=u\) for each value.
If \(u=-\frac{5\pi}{3}\), then
\(3\theta=-\frac{5\pi}{3}+\frac{\pi}{2}=-\frac{7\pi}{6}\), so \(\theta=-\frac{7\pi}{18}\).
If \(u=-\frac{\pi}{3}\), then
\(3\theta=-\frac{\pi}{3}+\frac{\pi}{2}=\frac{\pi}{6}\), so \(\theta=\frac{\pi}{18}\).
If \(u=\frac{\pi}{3}\), then
\(3\theta=\frac{\pi}{3}+\frac{\pi}{2}=\frac{5\pi}{6}\), so \(\theta=\frac{5\pi}{18}\).
Therefore the solutions are
\(\theta=-\frac{7\pi}{18},\quad \frac{\pi}{18},\quad \frac{5\pi}{18}.\)
This completes the solution and gives the required result.
