What would happen if you divide each side of the equation $\cot\beta\cos2\beta = 2\cot\beta$ by $\cot\beta$?

Question

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?

Answer 1

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.

Answer 2

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 $$\cos2\beta-2 =0\implies \cos2\beta=2\implies \beta \notin \mathbb{R}$$

Answer 3

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.

Answer 4

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 $$ \cot\beta=0\qquad\texttt{or}\qquad\cos2\beta-2=0 $$

Answer 5

Division is only possible if $\cot\beta$ is not equal to zero. I think you have got your answer.