We have the equation $\quad x^2 \Big(\sqrt{1-y^2} \sin (x)-\sin (4 x)\Big)+2 x y=0\quad$ for $\,\,\,x>0\,\,$ and $y$ is real. I want to solve this equation for $y$. Solving the quadratic equation, I obtain: $$y=\frac{\pm \sqrt{2} \sqrt{x^4 \sin ^2(x) (-\cos (2 x))+x^4 \sin ^2(x) \cos (8 x)+8 x^2 \sin ^2(x)}+4 x \sin (4 x)}{2 \left(x^2 \sin ^2(x)+4\right)}$$ Now, when I plot this function, I get
Can please someone explain why it always lays in $[-1,1]$? And, if it is true, then, is there a way to prove that the range of this function is $[-1,1]$?
If $|y| > 1$, how are you planning to get $x\left(\sqrt{1-y^2} \sin(x) - \sin(4x) \right)+2y$ to come out purely real?