Question

Given the derivative of f, f'(x)=sin(2x), state the critical values where f'(x)=0.

Answer 1

I substituted 0 for f'(x), such that \[0=\sin(2x)\]but I have no idea what I should do after that.

Answer 2

I mean I would know if it was sin(x) instead of sin(2x) but the 2x just throws me off.

Answer 3

Recall the trig identity: sin(2x) = 2sin(x)cos(x)

So: f'(x) = 2sin(x)cos(x)

Set that equal to zero and use it to find all of your critical points. (There are an infinite number without a given interval).

Answer 4

@Helloimjohn1234 Thank you!