Answer
$$\sin x=-\frac{\sqrt{6}}{6}$$
Work Step by Step
$$\cos2x=\frac{2}{3}\hspace{1.5cm}\pi\lt x\lt\frac{3\pi}{2}\hspace{1.5cm}\sin x=?$$
Though this exercise can certainly be solved using half-angle identities, if you do not want to deal with the square root, another way to do it is through double-angle identities.
- Double-angle Identity for cosines which only involves sines:
$$\cos2x=1-2\sin^2x$$
$$\sin^2x=\frac{1-\cos2x}{2}$$
$$\sin^2x=\frac{1-\frac{2}{3}}{2}$$
$$\sin^2x=\frac{\frac{1}{3}}{2}$$
$$\sin^2x=\frac{1}{6}$$
Thus, $$\sin x=\pm\frac{\sqrt1}{\sqrt{6}}=\pm\frac{1}{\sqrt6}=\pm\frac{\sqrt6}{6}$$
As $\pi\lt x\lt \frac{3\pi}{2}$, $x$ lies in the area of quadrant III, where sines are negative. So $\sin x\lt0$.
Therefore,
$$\sin x=-\frac{\sqrt{6}}{6}$$
