In the interval \([0,2\pi)\), you have \(\sin x=-1\) for \(x=\dfrac{3\pi}2\). Replacing \(x=\dfrac{2\theta}3\), you get
\(\dfrac{2\theta}3=\dfrac{3\pi}2\implies\theta=\dfrac32\times\dfrac{3\pi}2=\dfrac{9\pi}4\)
But \(2\pi=\dfrac{8\pi}4<\dfrac{9\pi}4\), so that solution for \(\theta\) falls outside the interval \([0,2\pi)\).
