Find $\sin \left(x\right)$ given $\cos \left(2x\right)=\frac{2}{3}$ and $\pi <x<\frac{3\pi }{2}$
I am trying to solve this problem by drawing a triangle in the appropriate quadrant. However, I am trying to drawn the triangle with angle $\cos 2x$ in quadrant III, as given by the criteria of the problem. However, $\cos 2x = 2/3$, but the triangle should be $-2/3$.
What is going on? Is it a double-angle issue, in that case should I draw the triangle in quadrant I or II?
$\cos 2x=\cos^2x-\sin^2x$ and $\sin^2x+\cos^2x=1$, then $$\cos 2x=1-2\sin^2x \Rightarrow \frac{2}{3}=1-2\sin^2x$$ $$\Rightarrow \frac{2}{3}-\frac{3}{3}=-2\sin^2x\Rightarrow \sin x=-\sqrt{\frac{1}{6}}=-\frac{1}{\sqrt{6}}. $$ The signal "$-$" is becouse $x\in 3^{\text{o}}$ quadrant and in this quadrant the sine is negative.