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i now know that its -sin(x)(2cos(x)+1)

can yu guys help me find the critical points

Are you sure it ask you to find f'(x) on [a,b]?

If so, you did that. You are done.

yes. may you help me find the critical points of this now

Set f'(x)=0 and solve for x to find the critical numbers.

You only need to solve this on the interval [0,2 pi]

can we do it together please

You are already factored f'(x)
Now set both factors equal to 0.

how?
can we work out the problem

step by step please

Yes. I'm walking you through it.
You have -sin(x)*(1+2cos(x))=0
Set both factors equal to 0.

ok

Just like if you have a*b=0 then either a=0 or b=0 or both=0

http://www.wolframalpha.com/input/?i=-sin%28x%29%282cos%28x%29%2B1%29%3D0

this is all i have

im confused about the results

Use your unit circle.
Keep in mind you only have to look between 0 and 2pi

-sinx=pin

\[\pi n\]