Answer: \(\alpha=-\dfrac{5\pi}{12},\ -\dfrac{\pi}{4},\ \dfrac{\pi}{12},\ \dfrac{\pi}{4}\).

We start with the main method. Factorise the trigonometric equation first, then solve each factor over the required interval.

Let

\(u=2\alpha-\frac{\pi}{3}.\)

The equation becomes

\(4\sin^2u=1,\)

so

\(\sin u=\pm\frac12.\)

Since

\(-\frac{\pi}{2}\leq \alpha\leq \frac{\pi}{2},\)

the corresponding range for \(u\) is

\(-\frac{4\pi}{3}\leq u\leq \frac{2\pi}{3}.\)

In this interval, the solutions for \(u\) are

\(-\frac{7\pi}{6},\quad -\frac{5\pi}{6},\quad -\frac{\pi}{6},\quad \frac{\pi}{6}.\)

Now use

\(2\alpha=u+\frac{\pi}{3}.\)

This gives

\(\alpha=-\frac{5\pi}{12},\quad -\frac{\pi}{4},\quad \frac{\pi}{12},\quad \frac{\pi}{4}.\)

This completes the solution and gives the required result.