# Math Help - trig derivative critical point

1. ## trig derivative critical point

find the critical points of $f(x)= 5sin^2x$ on $[0,\pi]$

$f'= 10sinxcosx$

I have a question when i get to this part.

$10sinxcosx=0$

When I set to zero, do I break into part then mutlipy the inverses of both sides?

10sin = 0 and cosx=0

2. ## Re: trig derivative critical point

Easier way (I think)

We have:
$5\sin{2x}=0$

This gives us that $2x=n\pi$ immediately, where n is an integer.

3. ## Re: trig derivative critical point

Originally Posted by delgeezee
find the critical points of $f(x)= 5sin^2x$ on $[0,\pi]$

$f'= 10sinxcosx$

I have a question when i get to this part.

$10sinxcosx=0$

When I set to zero, do I break into part then mutlipy the inverses of both sides?

10sin = 0 and cosx=0
You don't MULTIPLY the inverse on both sides, you just take the inverse...

But what you are saying is correct in essentials, so you now just need to find where sin(x) = 0 and cos(x) = 0.