#f(x) = 2sinx - cos^2x#
#rArr f'(x) = 2cosx - 2*cosx*sinx# (by the chain rule).
#= 2cosx - sin(2x)# (by the double-angle formula for #sin#).
#rArr f''(x) = -2sinx - 2cos(2x)#
#= -2(sinx - cos(2x))#
One of the conditions for a point of inflexion at #x=a# is that #f''(a) =0# ... however, #f''(x)# must also change sign.
First we solve the equation #f''(x) = 0#
#= -2(sinx - cos(2x)) = 0#
#rArr sinx - cos(2x) = 0#
Expanding #cos(2x)# with one of the double-angle formulas for #cos#:
#sinx - (1 - 2sin^2x) = 0#
#:. 2sin^2x + sinx - 1 = 0#
Now we let #u = sinx#
#2u^2 + u - 1 =0#
#rArr u = -1# or #u = 1/2# (by factorising the quadratic).
#rArr x in {pi/6, (5pi)/6, (3pi)/2}# (taking the inverse sine of the previous #u#-values in the given domain).
Now, to ensure that #f''(x)# changes sign across these #x#-values, we can just evaluate the following:
#f''(0) = 2#
#f''(pi/2) = - 4#
#f''(pi) =2#
#f''(2pi) = 2#
This shows that #f''(x)# changes sign across #x=pi/6# and #x=(5pi)/6# but not #x=(3pi)/2#, so the there is no point of inflexion at #x=(3pi)/2#
So, the co-ordinates of the points of inflexion for #f(x)# are:
#(pi/6, 1/4)# and #((5pi)/6, 1/4)#
Hope this helps!
