When is the function $f(x)$ equal to the second derivative of $f(x)$?

Question

Question: If $f(x) = \text{cot}(2x)$, $0 \le x \le 2\pi$ find all values of $x$ for which $f(x) = f''(x)$.

I was able to find the second derivative of $f(x)$ as $8[\text{csc}(2x)]^2[\text{cot}(2x)]$. However when I make the equation $\text{cot}(2x) = 8[\text{csc}(2x)]^2[\text{cot}(2x)]$ I cannot find a way to simplify the equation as cancelling any trigonometric functions changed the equation entirely. Using trigonometric identities didn't get me anywhere as I still ended up with trigonometric functions on both sides.

Answer 1

$8\csc^2(2x) \cot (2x) = \cot (2x)$ $8\csc^2(2x) \cot (2x) - \cot (2x)=0$ Factor $\cot 2x$

Answer 2

Hints:

• $f(x)=f(x)g(x) \implies f(x)=0 \text{ or } g(x)=1$

• $|\csc(y)| \ge 1$

• since $\cot(y)$ can be infinite, you may want to consider how to treat such a case