What would happen if you divide each side of the equation $\cot\beta\cos2\beta = 2\cot\beta$ by $\cot\beta$? 
Explain what would happen if you divide each side of the equation $\cot\beta \cos2\beta = 2\cot\beta$ by $\cot\beta$.
Is this a correct method to use when solving equations?

 A: You should avoid dividing where possible. Instead, rearrange: $$\begin{align}\cot\beta\cos(2\beta)-2\cot\beta&=0\\\cot\beta(\cos(2\beta)-2)&=0\\\implies\cot\beta=0\text{, or }\cos2\beta&=2\end{align}$$
Working in real angles $\beta$, the equation $\cos2\beta=2$ has no solutions, so your only option is that $\cot\beta=0\implies\beta=\pi/2+n\pi\text{ for }n\in\Bbb Z$.
This is a better method since if you divide, then you are implicitly assuming that $\cot\beta\neq0$, which could be the solution you are after.
A: So you have:
$$\cot\beta \cdot \cos(2\beta ) = 2\cdot \cot(\beta )$$
Dividing by $\cot\beta$, assuming that $\cot\beta\neq 0$ you get:
$$\frac{\cot\beta \cdot \cos(2\beta)}{\cot\beta} = \frac{2\cdot \cot(\beta )}{\cot\beta}$$
$$\implies \cos(2\beta) = 2$$
Avoiding assumptions about $\cot\beta \neq 0$, you can write it as:
$$\cot\beta \cdot \cos(2\beta ) - 2\cdot \cot(\beta ) = 0$$
$$\cot\beta(\cos2\beta - 2) = 0$$
Therefore, either $$\cot\beta =0 \implies \beta = \frac{\pi}{2} + n\pi$$ or $$\cos2\beta-2 =0\implies \cos2\beta=2\implies \beta \notin \mathbb{R}$$
A: It is valid, as long cotβ is not equal to 0 for your β.
If you're trying to find all solutions, you could do this for every β where cotβ is nonzero and check the other cases seperately.
A: No, it's not a correct method, because you discard solutions: indeed $\cot\beta$ could be zero.
The correct method is rewriting the equation as
$$
\cot\beta(\cos2\beta-2)=0
$$
which splits into
$$
\cot\beta=0\qquad\texttt{or}\qquad\cos2\beta-2=0
$$
A: Division is only possible if $\cot\beta$ is not equal to zero.
I think you have got your answer.
