f(x)=sin(π/2)cosx-sinxcos(π-x)的单调递增区间

Question

f(x)=sin(π/2)cosx-sinxcos(π-x)的单调递增区间

Answer 1

令f'(x)=-sinx+cos^2(x)-sin^2(x)=-sinx+1-2sin^2(x)=0得sinx=1/2或sinx=-1.
x=2k*pi+pi/3或2k*pi+2pi/3或x=2k*pi+3pi/2,
因为在[2k*pi+2pi/3,2k*pi+3pi/2]内导数大于零,故为单增区间