If $\frac{\sin2x}{\cos x}-1=0$, find $x$ given $\ \frac{\pi}{2}\le x\le\pi$ (cancelling out vs multiplying out the $\cos x$) To solve the equation
$$\frac{\sin2x}{\cos x}-1=0 \tag1$$
I manipulated it into the following:
$$\frac{2\sin x\cos x}{\cos x}-1=0 \tag2$$
Then, by multiplying $\cos x$ out, I obtained:
$$2\sin x\cos x-\cos x=0 \tag3$$
However, according to the answers I was supposed to cancel out the $\cos x$ which results in
$$2\sin x-1=0 \tag4$$ Thus, I myself attained the answers $\frac{5\pi}{6},\ \frac{\pi}{2}$, whilst the answer booklet merely had $\frac{5\pi}{6}$ .
I do not understand the logic behind cancelling out the $\cos x$ - I thought we were supposed to maximise the number of answers obtained by multiplying out rather than cancelling.
Any help clearing up my confusion would be much appreciated.

The confusion is due to the violation of equivalence conditions between mathematical steps.


Problem statement:
$$\frac{\sin2x}{\cos x}-1=0$$
where,  $\frac {\pi}{2}\leqslant x\leqslant\pi$.

$\rm Step-1.$
$$\frac{2\sin x\cos x}{\cos x}-1=0$$
There is no issue here. Because: $\sin 2x=2\sin x\cos x$.
$\rm Step-2.$
$$2\sin x\cos x-\cos x=0$$
We want to analyze this step.  When we multiply both side of the equation by $\cos x$, we should do this under the restriction $\cos x≠0$.  So why?

*

*No matter what mathematical operation we apply, before we start solving the equation $\frac{\sin2x}{\cos x}-1=0$, it is necessary to state that $\cos x≠0$.  Because, we can never make the denominator $0$.


*Applying the multiplication we have:
$$\cos x \left(\frac{2\sin x\cos x}{\cos x}-1\right)=0\cdot \cos x$$
This is essentially equivalent to:
$$\cos x \cdot \frac{2\sin x\cos x}{\cos x}-\cos x=0$$
Then, we can easily reduce the last equation to the following equivalent equation:
$$\begin{align}\frac {\cos x}{\cos x} \cdot \left(2\sin x\cos x\right)-\cos x=0\end{align}$$
Think about: if $\cos x=0$, can we write $\frac {\cos x}{\cos x}=\frac 00=1$ ?
Since $\cos x$ can not equal to $0$, then the multiplication under the restriction $\cos x≠0$, (even if $\cos x$ is not in the denominator) doesn't produce roots that are called extraneous roots in mathematics.

Therefore, the original equation $\frac{\sin2x}{\cos x}-1=0$ is equivalent to the derived equation $2\sin x\cos x-\cos x=0$, if and only if, when $\cos x≠0$.


The possible correct solution can be reached with the following $2$ ways:
$\require {cancel}$
$$\begin{align}&\frac{\sin2x}{\cos x}-1=0\\
\iff &\frac {2\sin x\cdot\cancel{ {\color{#c00}{\cos x}}}}{\cancel{\color{#c00}{{\cos x}}}}-1=0\\
\iff &2\sin x-1=0,\thinspace \cos x≠0\\
\iff &\sin x=\frac 12,\thinspace\cos x≠0\end{align}$$
or
$$\begin{align}&\frac{\sin2x}{\cos x}-1=0\\
\iff &\frac {2\sin x\cos x}{\cos x}-1=0\\
\iff &\frac {2\sin x\cos x-\cos x}{\cos x}=0.\end{align}$$
Then we know that,  $$\frac {A(x)}{B(x)}=0\iff A(x)=0\wedge B(x)≠0.$$
Therefore, we have:
$$2\sin x\cos x-\cos x=0,\thinspace\cos x≠0$$
This leads to:
$$\begin{align}&\cos x\left( 2\sin x-1\right)=0,\thinspace\cos x≠0\\
\iff &2\sin x-1=0,\thinspace\cos x≠0\\
\iff &\sin x=\frac 12,\thinspace \cos x≠0.\end{align}$$
Finally, under the restriction $\frac {\pi}{2}\leqslant x\leqslant\pi$, we obtain:
$$\sin x=\frac 12\wedge x≠\frac {\pi}{2}$$
Thus, indeed the answer is $x=\frac {5\pi}{6}$.
