<s>[INST] 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 $x\to \infty$.

*

*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.

 [/INST] 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 $3 x^2-5 = 0$).
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