To find the inflection points we have to calculate the second derivative then set the equation to zero.
Knowing these properties:
#color(blue)((sinu)'=u'cosu#
#color(red)((cosu)'=-u'sinu#
#color(green)(v^n=nv^(n-1))#
#color(orange)(sin2x=2sinxcosx#
First let us find f''(x):
#f'(x)=color(green)(2sinx(sinx)'-2(cosx)(cosx)')# using 3rd property
#f'(x)=2sinxcolor(blue)(cosx)-2cosx(color(red)(-sinx))#
#f'(x)=2sinxcosx+2cosxsinx#
#f'(x)=4sinxcosx#
#f'(x)=2color(orange)((2sinxcosx))#
#f'(x)=color(orange)(2sin2x)#
Let us find #f''(x)#:
#f''(x)=(f'(x))'#
#f''(x)=(2sin2x)'#
#f''(x)=2color(blue)((sin2x))'#
#f"(x)=2*(color(blue)(2cos2x))# using first property above
#f"(x)=4cos2x#
Now set #f"(x)=0#
#4cos2x=0#
#cos2x=0#
In the given interval #|0,2pi|# then :
#2x=(pi)/2rArrx=(pi)/4#
#f((pi)/4)=sin^2(pi/4)-cos^2(pi/4)=((sqrt2)/2)^2-(sqrt2/2)^2=0#
The point is#(color(brown)(pi/4,0))#
Or
#2x=(3pi)/2rArrx=(3pi)/4#
#f((3pi)/4)=sin^2((3pi)/4)-cos^2((3pi)/4)#
#f((3pi)/4)=(sin(pi-(pi)/4))^2-(cos(pi-(pi)/4))^2#
#f((3pi)/4)=(sin(pi/4))^2-(-cos(pi/4))^2#
#f((3pi)/4)=((sqrt2)/2)^2-((sqrt2)/2)^2=0#
The second point is #(color(brown)((3pi)/4,0))#
