Why does cancelling by $\sin x$ when solving $4\tan x = 5\sin x$ for $0\leq x < 2\pi $ miss solutions? So the solution to the problem is:
$$4 \tan x = 5 \sin x$$
$$4 \frac{\sin x}{\cos x} = 5 \sin x$$
$$4 \sin x = 5 \sin x \cos x$$
So, either $\sin x = 0$ or $\cos x = 4/5$. I'm fine with the logic of this. But is there some deeper reason why cancelling the $\sin x$ straight away loses solutions?
 A: For the same reason why solving $x = x^2$ by dividing by $x$ would lose solutions.
If you divide both sides by something, you implicitly give rise to two cases.
For $x=x^2$, if you divide by $x$, then you get $1=x$, but only if $x \ne 0$. If $x$ (what you divided by) is equal to zero, then you just divided by zero, and hence have to treat that case separately.
That is,
$$x = x^2 \implies \underbrace{\frac{x}{x} = \frac{x^2}{x}}_{1 = x, \text{ if } x \ne 0} \text{ or } x = 0$$
In your scenario,
$$4 \sin x = 5 \cos x \sin x 
\implies \underbrace{\frac{4 \sin x}{\sin x} = \frac{5 \cos x \sin x}{\sin x}}_{4 = 5 \cos x\text{ if } \sin x \ne 0} \text{ or } \sin x = 0$$
Sometimes that equals-to-zero case gives rise to other solutions, sometimes not, it just depends. Here, it does, because if $\sin x = 0$, then both sides of the equation are equal to zero, so what remains is to find the $x$ that work.
A: We can’t cancel terms on both sides without knowing whether it is zero.
Just like solving
$x^2=x \Leftrightarrow  x^2-x=0 \Leftrightarrow x(x-1)=0 \Leftrightarrow x=0 \textrm{ or }x=1, \tag*{} $
we shall miss the solution $x=0.$
Back to our equation, we have
$$
\begin{aligned}
& 4 \tan x=5 \sin x \\
\Leftrightarrow \quad  & \frac{4 \sin x}{\cos x}=5 \sin x \\
\Leftrightarrow \quad& 4 \sin x=5 \sin x \cos x  \\
\Leftrightarrow \quad& \sin x(5 \cos x-4)=0 \\
\Leftrightarrow \quad& \sin x=0 \text { or } 5 \cos x-4=0 \\
\Leftrightarrow \quad& x=n \pi \quad \text { or }\quad  \cos x=\frac{4}{5}\\
\Leftrightarrow  \quad & x=n \pi \quad  \textrm{ or } \quad  2 n \pi \pm \cos ^{-1}\left(\frac{4}{5}\right)
\end{aligned}
$$
where $n\in Z$.
If we cancel $\sin x$ on both sides, we shall miss the solutions $x=n\pi$.
