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Answer 1 · 2014-11-08T00:57:59

By taking the derivative,

#f'(theta)=-2sin theta+cos2theta cdot2#

by factoring out #-2#,

#=-2(sin theta-cos2theta)#

by #cos2theta=1-2sin^2theta#,

#=-2(2sin^2theta+sin theta-1)#

by factoring out,

#=-2(2sin theta-1)(sin theta+1)=0#

#=>{(sin theta=1/2 => theta=pi/6","{5pi}/6),(sin theta=-1 => theta={3pi}/2):}#

Hence, critical numbers are #theta=pi/6, {5pi}/6, {3pi}/2#.

I hope that this was helpful.