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Câu 2. A. Ta có $\sin 2x=2\sin x\cos x$. Vì $\pi< x< \frac{3\pi}{2}$ nên $\cos x< 0$. Từ $\sin^2 x + \cos^2 x = 1$, ta có $\cos x = -\sqrt{1-\sin^2 x} = -\sqrt{1-\left(-\frac{2}{3}\right)^2} = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}$. Do đó, $\sin 2x = 2 \cdot \left(-\frac{2}{3}\right) \cdot \left(-\frac{\sqrt{5}}{3}\right) = \frac{4\sqrt{5}}{9}$. B. Ta có $\sin x + \cos x = \frac{1}{3}$. Vì $\sin^2 x + \cos^2 x = 1$, ta có $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x$. Do đó, $\left(\frac{1}{3}\right)^2 = 1 + 2\sin x \cos x$, suy ra $\frac{1}{9} = 1 + 2\sin x \cos x$, suy ra $2\sin x \cos x = \frac{1}{9} - 1 = -\frac{8}{9}$. Ta có $\cos 2x = \cos^2 x - \sin^2 x = (\cos x + \sin x)(\cos x - \sin x)$. Vì $\sin x + \cos x = \frac{1}{3}$, ta có $\cos x - \sin x = \sqrt{(\cos x - \sin x)^2} = \sqrt{1 - 2\sin x \cos x} = \sqrt{1 - \left(-\frac{8}{9}\right)} = \sqrt{\frac{17}{9}} = \frac{\sqrt{17}}{3}$. Do đó, $\cos 2x = \left(\frac{1}{3}\right) \cdot \left(\frac{\sqrt{17}}{3}\right) = \frac{\sqrt{17}}{9}$. Cuối cùng, $\cos 4x = 2\cos^2 2x - 1 = 2\left(\frac{\sqrt{17}}{9}\right)^2 - 1 = 2 \cdot \frac{17}{81} - 1 = \frac{34}{81} - 1 = \frac{34}{81} - \frac{81}{81} = -\frac{47}{81}$.