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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.
Just like if you have a*b=0 then either a=0 or b=0 or both=0
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
You have -sin(x)=0 or 1+2cos(x)=0 Solve both of these for the trig function. Solve the first one for sin(x) and solve the second one for cos(x). Tell me what you get after doing this.
Well you are trying to solve -sin(x)=0 for x. You could divide both sides by -1 since -1 will never be 0. Try solving this: sin(x)=0 Then we will come back to 1+2cos(x)=0