Answer: \(\theta=-\dfrac{\pi}{3},\,0,\,\dfrac{\pi}{3}\).

Start with

\(2\sin^3\theta=3\sin\theta\cos\theta.\)

Bring all terms to one side:

\(2\sin^3\theta-3\sin\theta\cos\theta=0.\)

Factor out \(\sin\theta\):

\(\sin\theta(2\sin^2\theta-3\cos\theta)=0.\)

Therefore either

\(\sin\theta=0\)

or

\(2\sin^2\theta-3\cos\theta=0.\)

From \(\sin\theta=0\), in the interval

\(-\dfrac\pi2\leqslant\theta\leqslant\dfrac\pi2,\)

we get

\(\theta=0.\)

For the second factor, use \(\sin^2\theta=1-\cos^2\theta\):

\(2(1-\cos^2\theta)-3\cos\theta=0.\)

Expanding,

\(2-2\cos^2\theta-3\cos\theta=0.\)

Multiply by \(-1\):

\(2\cos^2\theta+3\cos\theta-2=0.\)

Factorise:

\((2\cos\theta-1)(\cos\theta+2)=0.\)

The equation \(\cos\theta=-2\) has no solution, so

\(\cos\theta=\dfrac12.\)

In the interval \(-\dfrac\pi2\leqslant\theta\leqslant\dfrac\pi2\), this gives

\(\theta=-\dfrac\pi3\quad\text{or}\quad\theta=\dfrac\pi3.\)

Including the solution from \(\sin\theta=0\), the solutions are

\(\theta=-\dfrac\pi3,\quad0,\quad\dfrac\pi3.\)