Finding the critical points of the damped oscillation equation $f(x) = e^{-x} \sin 2 x$ I have the following function:
$$ f(x) = e^{-x} \sin(2x) $$
I have found the derivative of the function to be
$$ f'(x) = e^{-x} (2\cos(2x)-\sin(2x)) .$$
How would one go about solving $ f'(x) = 0 $?
I have tried multiple ways of going about it, but I don't seem to get it quite right.
Range: $ x\in [0, 2\pi] $

Hint Since $e^{-x}$ is never zero, $f'(x) = 0$ iff $$2 \cos 2x - \sin 2x = 0 .$$
Can we rearrange this equation to write it in terms of a single trigonometric function?
