Why solutions of $x^2 \left(\sqrt{1-y^2} \sin (x)-\sin (4 x)\right)+2 x y=0$ for $y$ are not verified? I have this equation
$$x^2 \left(\sqrt{1-y^2} \sin (x)-\sin (4 x)\right)+2 x y=0$$
for $x>0$ and $y$ real.
Solving this equation for $y$, by Mathematica, I obtain these two solutions:
$$y=\frac{\pm \sqrt{2} \sqrt{x^4 \sin ^2(x) (-\cos (2 x))+x^4 \sin ^2(x) \cos (8 x)+8 x^2 \sin ^2(x)}+4 x \sin (4 x)}{2 \left(x^2 \sin ^2(x)+4\right)}$$
But, then when I substitute them in the original equation, it is not verified. Where I am doing wrong? And, if these solutions are wrong, how can I solve this equation to obtain $y$?
Any comment is welcome.
 A: Here is a form of solution that will make you understand why you have "extraneous" roots.
Simplifying by $x$ (possible because $x > 0$) and grouping the two terms having a $y$ on a same side:
$$\sqrt{1-y^2}x \sin x+2y=x \sin 4x$$
If you put to the square both sides (knowing that in this way you introduce artificial roots), you get:
$$(1-y^2)(x \sin x)^2+4y\sqrt{1-y^2}(x \sin x)+4y^2=x \sin 4x$$
Said otherwise, moving some terms RHS/LHS:
$$(1-y^2)(x \sin x)^2+4y^2-x \sin 4x= -4y\sqrt{1-y^2}(x \sin x)$$
Squaring once again (possibly introducing again superfluous roots), one gets a 4th degree equation in $y$, which in fact contains only terms in $Y=y^2$ ; therefore one gets a quadratic equation in variable $Y$.
Finally, when returning to $y$ variable, one should obtain solution(s) that should the same as the ones found by Mathematica (one can indeed observe, without having done the final calculation the characteristic form $\dfrac{-b\pm\sqrt{d}}{2a}$ of the solutions to a quadratic equation) with the same drawback of the presence of extraneous roots.
But, as said upwards, all the roots will not be roots of the initial equation (broadly saying, we have multiplied by 4 the number of roots).
Remark: the introduction of fictitious roots is explainable on a naive simple example: starting from equation $x=1$ which has a unique root (evidently), squaring it gives $x^2=1$ with two roots $x=\pm 1$.
A: Let the function $y$ returned by Mathematica for the positive root be denoted as $y_+(x)$, and $y_-(x)$ denote the negative root. The solution(s) in $y$ for a given $x$ for $$x^{2}\left(\sqrt{1-y^{2}}\sin\left(x\right)-\sin\left(4x\right)\right)+2xy=0$$
will be in $\{y_-(x), y_+(x)\}$. However, the problem is that the Mathematica solutions may also "cover" regions where it doesn't line up with the original equation. If Mathematica was completely perfect, it would also have indicated the intervals where each solution $\left(y_-(x) \text{ and } y_+(x)\right)$ was an actual solution to the equation.
