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Mathematics 21 Online
OpenStudy (anonymous):

sin^2x+cosx 0

OpenStudy (anonymous):

Wait! Do you have to find the derivative first? cos^2x-sinx

OpenStudy (anonymous):

For critical numbers, begin by differentiating, to obtain 2sin(x)*cos(x)-sin(x)=0 solve for x.

OpenStudy (anonymous):

2sin(pi)*cos(pi)-sin(pi)=0?

OpenStudy (anonymous):

yes, sin(pi)=0 so 2sin(pi)cos(pi)-sin(pi)=0, but 0 isn't within the bounds try x=1.04.

OpenStudy (anonymous):

more accurately x= 1.047197

OpenStudy (anonymous):

2sin(pi/3)cos(pi/3)-sin(pi/3)=0

OpenStudy (anonymous):

2sin(5pi/3)cos(5pi/3)-sin(5pi/3)=0

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