Great Radical Equation

Let $x=\sin \theta$ as suggested, then:

$\begin{aligned} 4x\sqrt{1-x^2}-\sqrt{2} & = 0 \\ 4x\sqrt{1-x^2} & = \sqrt{2} \\ 4\sin \theta \sqrt{1-\sin^2 \theta} & = \sqrt{2} \\ 4 \sin \theta \cos \theta & = \sqrt{2} \\ 2 \sin (2\theta) & = \sqrt{2} \\ \sin (2\theta) & = \frac{1}{\sqrt{2}} \\ \Rightarrow 2 \theta & = \frac{\pi}{4} \\ \theta & = \frac{\pi}{8} \quad \quad \small \color{#3D99F6}{\text{for } 0 \le \theta \le \frac{\pi}{2}} \end{aligned}$

$\Rightarrow b - a = 8 - 1 = \boxed{7}$

note that x must be positive. now solve $4x\sqrt{1-x^2}=\sqrt{2}$ $16x^2-16x^4=2$ $8x^4-8x^2+1=0$ $x^2=\dfrac{2\pm\sqrt{2}}{4}$ $x=\dfrac{\sqrt{2\pm\sqrt{2}}}{2}$ negative neglected. since in the first quadrant sin (y)<cos (y), $cos (y) =\dfrac{\sqrt{2+\sqrt{2}}}{2},sin (y) =\dfrac{\sqrt{2-\sqrt{2}}}{2}$ $cos (2y)=cos^2(y)-sin^2(y)=\dfrac{2+\sqrt{2}}{4}-\dfrac{2-\sqrt{2}}{4}=\dfrac{\sqrt{2}}{2}$ $2y=\dfrac{\pi}{4}\Longrightarrow y=\dfrac{1\pi}{8}$ $8-1=\boxed{7}$ not that other solutions to 2y neglected as conditions for first quadrant.

$x\in [-1,1]\quad hints\quad us\quad to\quad substitute\quad x=\quad cos\theta .\\ \sqrt { 1-{ x }^{ 2 } } =\sqrt { 1-{ cos }^{ 2 }\theta } \\ \quad \quad \quad \quad \quad \quad =|sin\theta |\\ since\quad \theta \in \quad 1st\quad quadrant\\ |sin\theta |=sin\theta \\ \therefore \quad 4cos\theta sin\theta \quad =\sqrt { 2 } \\ sin2\theta =\frac { 1 }{ \sqrt { 2 } } \\ sin2\theta =sin\frac { \pi }{ 4 } \\ \theta =\frac { \pi }{ 8 } \\ x=cos\frac { \pi }{ 8 } \\ a=1,b=8\\ b-a=\boxed { 7 } .$