Allowed values of $x$ in $x \sin 6 y-\left(3 x^2-5\right) \sin ^23 y=0$ when $y=0$? Can someone please check the given explanation? I have this two-variable equation
$$x \sin 6 y-\left(3 x^2-5\right) \sin ^23 y=0$$
for $0\leqslant x,y<\infty$, and I want to check which values of $x$ satisfy this equation when $y=0$.
From the above equation, I see that when $y=0$, it is fulfilled for all  $0\leqslant x<\infty$.
Then, I rewrite the equation as
$$\frac{x}{3 x^2-5}=\frac{1}{2} \tan (3 y),$$
from here, I see that when $y=0$, there are only two solutions for $x$, i.e. $x=0$ and $x\to \infty$.

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*Then, what can I say in general about the solutions of $x$ when $y$ is zero?

*And what can we say about the point $x=\sqrt{\frac{3}{5}}$? Is it an acceptable solution for $3y=\frac{\pi }{2}$? Since from the first equation, it can be a solution, but I have doubts about it from the second form of the equation.


When you "rewrite" an equation by multiplying or dividing by some factor, the new equation is equivalent to the old when that factor is not $0$.  When it is $0$, the two are generally not equivalent.
So, your first equation
$$ x \sin 6 y - (3 x^2 - 5) \sin^2 3 y $$
is indeed always true when $y = 0$, or more generally when $\sin 3 y = 0$.  When you divide by $(3 x^2 - 5) \sin 6 y$ to get your second equation
$$ \frac{x}{3x^2-5} = \frac{1}{2} \tan 3y $$
this is equivalent to the first as long as $3 x^2 - 5 \ne 0$ and $\sin 6 y \ne 0$.  When
$\sin 3y = 0$, the right side is $0$ but the left is not (unless $x=0$).  When $3x^2 - 5 = 0$ the left side is undefined but the right side is defined (unless $\cos 3 y = 0$).
When $\cos 3y = 0$ the right side is undefined but the left side is defined (unless $3 x^2-5 = 0$).
