Clearly $ -1\le x\le 1$.so we can let $ x=\cos{y}(0\le y\le \pi)$,then $ \sqrt{1-x}=\sqrt{1-\cos{y}}=\sqrt{2\sin^2{y/2}}=\sqrt{2}\sin(y/2)$, $ 2x^2-1=2\cos^2{y}-1=\cos{2y}$,and $ 2x\sqrt{1-x^2}=2\cos{y}\sin{y}=\sin{2y}$, so we have the equation is equivalent to $ \sqrt{2}\sin(y/2)=\cos{2y}+\sin{2y}=\sqrt{2}\sin(2y+\pi /4)$ $ \Longleftrightarrow \sin(y/2)=\sin(2y+\pi /4)$. so $ 2y+\pi /4=y/2+2m\pi,2y+\pi /4=(\pi-y/2)+2m\pi(m\in \mathbb{Z})$,because $ 0\le y\le \pi$,we have $ y=3\pi /10$, thusthe root is $ x=\cos(3\pi /10)=\frac{\sqrt{10-2\sqrt{5}}}{4}$.